{"id":13940,"date":"2019-12-21T16:22:52","date_gmt":"2019-12-21T10:52:52","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13940"},"modified":"2021-07-02T15:35:39","modified_gmt":"2021-07-02T10:05:39","slug":"1st-puc-maths-question-bank-chapter-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-maths-question-bank-chapter-7\/","title":{"rendered":"1st PUC Maths Question Bank Chapter 7 Permutations and Combinations"},"content":{"rendered":"

Students can Download Maths Chapter 7 Permutations and Combinations Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Maths Question Bank Chapter 7 Permutations and Combinations<\/h2>\n

Question 1.
\nState fundamental principle of counting (or multiplication principle) for
\n(i) two events,
\n(ii) three events.
\nAnswer :
\n(i) Fundamental principle for two events states that \u201cIf an event can occur in m different ways, following which another even occur in n different ways, then the total number of occurrence of the events in the given order is m x n.<\/p>\n

(ii) Fundamental principle for three events states that, \u201cIf an event can occur in m different ways, following which second event occurs in n different ways, following third event can occur in p different ways, then the total number of occurrence of the events in the given order is<\/p>\n

Question 2.
\nHow many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
\n(i) Repetition of the digits is allowed?
\n(ii) Repetition of the digits is not allowed?
\nAnswer :
\nGiven digits are 1, 2, 3, 4, 5.
\nTo find : number of 3-digit numbers,
\n(i) Repetition of digit is allowed.
\n\"1st
\nSince repetition of digits is allowed, then each place can be filled by any of 5 given digits in 5 ways. Hence, three places can be filled in 5 x 5 x 5 ways i.e., 125 ways (By fundamental principle of counting). Required number of numbers = 125<\/p>\n

(ii) Repetition of digit is not allowed
\n\"1st<\/p>\n

Since the repetition of digits is not allowed then hundred\u2019s place can be filled by any one of 5 given digits in 5 ways, ten\u2019s place can be filled by any one of 4 left digits in 4 ways and unit\u2019s place can be filled by any one of 3 left digits in 3 ways.
\nHence, three places can be filled in
\n5 x 4 x 3 ways i.e., 60 ways.
\n\u2234 Required number of numbers = 60<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nHow many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
\nAnswer :
\nGiven digits are 1, 2, 3, 4, 5, 6
\nTo find : number of 3-digit even numbers
\n\"1st<\/p>\n

Since the required number is even, it has to be end with 2, 4 or 6, therefore unit\u2019s place can be filled in only 3 ways.<\/p>\n

Since the repetition of the digits is allowed, therefore hundred\u2019s place and ten\u2019s place can be filled each in 6 ways.<\/p>\n

Hence three places can be filled in 6 x 6 x 3 ways i.e., 108 ways.
\n\u2234 Required number of numbers = 108<\/p>\n

Question 4.
\nHow many 4-letter code can be formed using the first 10 letters of English alphabet, if no letter can be repeated?
\nAnswer :
\nGiven : First 10 letters of English alphabet
\nTo find : number of 4-letter codes
\n\"1st<\/p>\n

Since required is 4-letter code, therefore, I-place can be filled in 10 ways, H, III and IV places can be filled in 9 ways, 8 ways and 7 ways respectively (because repetition of letter is not allowed).<\/p>\n

Hence four places can be filled in
\n10 x 9 x 8 x 7 ways
\ni.e., 5040 ways
\n\u2234 Required number of 4-letter code = 5040<\/p>\n

Question 5.
\nHow many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
\nAnswer :
\nGiven : 0 to 9 digits (i.e., 10 digits)
\nTo find : number of 5-digit telephone numbers
\n\"1st
\nThere are 10 digits form 0 to 9 out of which 6 and 7 are fixed in first and second places respectively. Since the repetition of digits is not allowed, therefore the third, fourth and fifth places can be filled in 8, 7 and 6 ways respectively.<\/p>\n

Hence five places can be filled in 8 x 7 x 6 ways i.e., 336 ways
\nRequired number of telephone numbers = 336<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA coin is tossed 3 times and the out comes are recorded. How many possible outcomes are there?
\nAnswer :
\nWhen a coin is tossed once, there are two possible out comes head or tail when a coin is tossed three times, each time there are two possible out comes
\nHence total number of out comes =2 x 2 x 2 = 8 .
\nNote: Possible out comes are {HHH, HHT, HTH, HIT, THH, THT, TTH, TIT}<\/p>\n

Question 7.
\nGiven 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
\nAnswer :
\nGiven : 5 different colour flags.
\nTo find: Number of different signals, using 2 flags
\n\"1st
\nWe have to select two flags one after the other. Therefore the upper flag (first flag) can be selected in 5 ways. Since both flags must be different colour, lower flag (second flag) can be selected in 4 ways.
\n\u2234 Number of different signals = 5 x 4 = 20<\/p>\n

Question 8.
\nFind the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
\nAnswer :
\nGiven work is ROSE
\nTo find : Number of 4 letter words
\n\"1st
\nSince required is 4 letter word in which repetition of letter is not allowed, therefore first place can be filled in 4 different ways and second, third and fourth places can be filled in 3 ways, 2 ways and 1 way respectively.
\nHence four places can be filled in 4 x 3 x 2 x 1 ways
\ni.e., 24 ways
\n\u2234 Required number of words = 24<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nGiven 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other.
\nAnswer :
\nGiven : 4 flags of different colours
\nTo find : Number of different signals, using 2 flags
\n\"1st
\nTherefore the first flag can be selected in 4 ways.
\nSince both flags must be different colour, second flag can be selected in 3 ways . Number of different signals = 4 x 3 = 12<\/p>\n

Question 10.
\nHow many 2-digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits, can be repeated?
\nAnswer :
\nGiven : 1, 2, 3, 4, 5 digits (5 digits)
\nTo find : 2-digit even numbers
\n\"1st
\n\u2234 Required 2- digits even number
\n= 5 x 2 = 10<\/p>\n

Question 11.
\nFind the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.
\nAnswer :
\nGiven : 5 different flags.
\nTo find : Number of different signals
\nSignals that can be generated by arranging at least 2 flags means a signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags.<\/p>\n

Case (I):
\nSignals consisting of 2 flags
\n\"1st<\/p>\n

We have to choose two flags one after the other. Therefore the first place can be filled in 5 ways and then the second place can be filled in 4 ways. Number of different signals that can be formed = 5 x 4 = 20<\/p>\n

Case (II):
\nSignals consists of 3 flags
\n\"1st
\n\u2234 Required number of different signals can be formed = 5 x 4 x 3 x = 60<\/p>\n

Case (III):
\nSignals consisting of 4 flags
\n\"1st<\/p>\n

\u2234 Required number of different signals can be formed = 5 x 4 x 3 x = 120<\/p>\n

Case (IV):
\nSignals consisting of 5 flags
\n\"1st
\n\u2234 Required number of different signals can be formed = 5 x 4 x 3 x 2 x 1= 120
\nTherefore,total number of signals = 20 + 60 + 120 + 120 = 320<\/p>\n

Permutations:<\/p>\n

Question 12.
\nDefine permutations.
\nAnswer :
\nPermutation is an arrangement in definite order of a number of objects taken some or all at a time.<\/p>\n

Observations:<\/strong><\/p>\n

1. The number of permutations of n different objects taken r at a time, where 0<r<n and the objects do not repeat is
\nn(n-1)(n – 2)…… (n – r +1)
\nwhich is denoted by n<\/sup>Pr
\n<\/sub>i.e., n<\/sup>Pr<\/sub>= n(n – 1)(n – 2)……… (n – r +1)<\/p>\n

2. The number of permutations of n different objects taken r at a time where repetition is allowed in nr<\/sup>.<\/p>\n

3. Factorial rotation :
\nThe notation n! represents the product of first n natural numbers.
\ni.e., n! = 1 x 2 x 3 x…… .x(n-1) x n<\/p>\n

Remark:<\/p>\n

(i) 0! = 1,1 x 2 = 2!, 1 x 2 x 3 = 3!
\nAnd so on.<\/p>\n

(ii) n! = n(n-1)!
\n= n(n – 1)(n – 2)! (n\u22652)
\n= n(n – 1)(n – 2)(n – 3)! (n\u22653)<\/p>\n

4. Formula for n<\/sup>Pr<\/sub>
\n\\(P_{r}=\\frac{n !}{(n-r) !}, 0 \\leq r \\leq n\\)
\nNote : n<\/sup>Pn<\/sub> =n! and n<\/sup>Pa<\/sub> =1<\/p>\n

5. The number of permutations of n objects, where p objects are of the same kind and rest are all different is equal to<\/p>\n

6. The number of permutations of n objects, where px<\/sub> objects are of one kind, p2<\/sub> are of second kind,….. pk<\/sub> are of kth<\/sup> kind and the rest, if any, are of different kind is
\n\\(\\frac{n !}{p_{1} ! p_{2} ! \\ldots \\ldots p_{k} !}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nEvaluate:
\n(i) 5!
\n(ii) 7!
\n(iii) 8!
\n(iv) 4! – 3!
\n(v) 7! – 5!
\nAns:
\n(i) 5!\u00a0 = 5 x 4 x 3 x 2 x 1 = 120
\n(ii) 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
\n(iii) 8!\u00a0 = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
\n(iv) 4! – 3! = (4 x 3 x 2 x 1) – (3 x 2 x 1) = 24-6= 18
\n(v) 7! – 5! = (1 x 2 x 3 x 4 x 5 x 6 x 7) -(1 x 2 x 3 x 4 x 5) = 5040 – 120 = 4920<\/p>\n

Question 14.
\nCompute:
\n(i)\\(\\frac{7 !}{5 !} \\)
\n(ii)\\(\\frac{8 !}{6 ! \\times 2 !}\\)
\n(iii) \\(\\frac{12 !}{10 ! \\times 2 !}\\)
\nAnswer:
\n\"1st<\/p>\n

Question 15.
\nFind x,if
\n(i) \\(\\frac{1}{6 !}+\\frac{1}{7 !}=\\frac{x}{8 !}\\)
\n(ii)\\(\\frac{1}{8 !}+\\frac{1}{9 !}=\\frac{x}{10 !}\\)
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 16.
\nEvaluate :
\n\\(\\frac{n !}{(n-r) !}\\), when
\n(i) n=5,r=2
\n(ii) n=6,r=2
\n(iii) n=9,r=5
\nAnswer:
\n\"1st<\/p>\n

Question 17.
\nEvaluate :
\n\\(\\frac{n !}{r !(n-r) !}\\),When n=5, r=2.
\nAnswer:
\n\"1st<\/p>\n

Question 18.
\nFind n, if
\n(i) \\( P_{s}=42^{n} P_{3}, n>4\\)
\n(ii) \\(\\frac{^{n} P_{4}}{^{n-1} P_{4}}=\\frac{5}{3}, n>4\\)
\n(iii) \\(^{n-1} P_{3}:^{n} P_{4}=1: 9\\)
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 19.
\nFind r, if
\n(i) \\(5 \\cdot^{4} P_{r}=6 \\cdot^{5} P_{r-1}\\)
\n(ii) \\(^{5} P_{r}=2 \\cdot^{6} P_{r-1}\\)
\n(iii) \\(^{5} P_{r}=^{6} P_{r-1}\\)
\nAnswer:
\n\"1st
\n\"1st
\n\"1st<\/p>\n

Question 20.
\nHow many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
\nAnswer :
\nThe required number of numbers is equal to the number of ways in which 9 digits (all different) can be arranged taken 3 at a time. This can be done in 9<\/sup>P3<\/sub> ways.
\n\u2234 Required number of numbers = \u00a09<\/sup>P3<\/sub>
\n= 9 x 8 x 7 = 504<\/p>\n

\"KSEEB<\/p>\n

Question 21.
\nHow many four-digits numbers are there with no digit repeated?
\nAnswer :
\nGiven : 0 to 9 digits (10 digits)
\nTo find : 4-digit numbers
\n\"1st
\nFirst, thousand\u2019s place on be filled by 9 ways from any one of the 1 to 9 digits (v \u20180\u2019 (zero) is not placed in thousand\u2019s place).
\nThen the remaining three places can be filled with the remaining 9 digits in 9<\/sup>P3<\/sub> ways.
\n\u2234 Required number of numbers = 9 x 9<\/sup>P3
\n<\/sub>= 9 x (9 x 8 x 7)
\n= 4536<\/p>\n

Question 22.
\nHow many three-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?
\nAnswer:
\nGiven : 1, 2, 3, 4, 6, 7 digits.
\nTo find : 3-digit even numbers
\n\"1st
\nSince required number is 3-digit even number, unit\u2019s place can be filled in 3 ways from any one of the even digits 2, 4, 6. Then the remaining two places can be filled in 5<\/sup> P2<\/sub> ways.
\n\u2234 Required = 5<\/sup>P2<\/sub> x 3 = (5 x 4) x 3 = 60<\/p>\n

Question 23.
\nFind the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
\nAnswer :
\nGiven : Digits 1, 2, 3, 4, 5 (5 digits)
\nTo find :
\n(i) 4-digits numbers
\n(ii) 4-digit event numbers
\n\"1st<\/p>\n

The required number of numbers is equal to the number of ways in which 5 digits can be arranged taken 4 at a time. This can be done in S<\/sup>P4<\/sub> ways.
\n\u2234 Required 4-digit n umber of numbers = 5<\/sup>P4<\/sub> =5 x 4 x 3 x 2 = 120<\/p>\n

(ii) To find the number of even numbers :
\n\"1st
\nIn the given digits, even digits are 2 and 4 Since required word is 8 letters word, with or without meaning, therefore these 8 letters can be arranged in 4<\/sup>P3<\/sub> ways.
\n\u2234 Required 4- digit number of numbers
\n2 x 4<\/sup>P3<\/sub> =2 x (4 x 3 x 2) = 48<\/p>\n

Question 24.
\nFrom a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
\nAnswer :
\nGiven : Number of persons in committee = 8
\nTo find : Two positions (chairman, vice chairman)
\n\"1st
\nSince one person cannot hold more than one position, therefore chairman post can be filled in 8<\/sup>P1<\/sub> ways and vice chairman post can be filled in 7<\/sup>P1<\/sub> ways.
\n\u2234 Required number of ways of choices =8<\/sup>P1 <\/sub>x 7<\/sup>P1
\n<\/sub>= 8 x 7 = 56<\/p>\n

Question 25.
\nHow many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
\nAnswer :
\nGiven : Work EQUATION (8 letters word)
\n\"1st
\nSince required word is 8 letters word, with or without meaning, therefore these 8 letters can be arranged in 8<\/sup>P8<\/sub> ways.
\n\u2234 Required number of words
\n= 8<\/sup>P8 <\/sub>= 8! = 40320<\/p>\n

\"KSEEB<\/p>\n

Question 26.
\nHow many words, with or without meaning can be made form the letters of the word MONDAY,assuming that no letter is repeated, if
\n(i) 4 letters are used at a time,
\n(ii)\u00a0all letters are used at a time,
\n(iii) all letters are used but first letter is vowel?
\nAnswer:
\nGiven work MONDAY has six different letters
\n(i) When only 4 letters are to be used, then the number of words formed
\n= 6 x 5 x 4 x 3 = 360<\/p>\n

(ii) When all letters are to be used, then the number of words formed
\n= 6<\/sup>P6<\/sub> =61 = 6 x 5 x 4 x 3 x 2 x 1 = 720<\/p>\n

(iii) In the given word, vowels are A, O
\n\"1st
\nSince there are only two vowels A and O, therefore the first place can be filled in 2<\/sup>P1<\/sub>
\n(2 ways) ways and the remaining 5 places can be filled from remaining 5 letters,
\nin\u00a0 5<\/sup>P5 <\/sub>ways.
\n\u2234 Required number of words
\n= 2<\/sup>P1 <\/sub>x 5<\/sup>P5 <\/sub>
\n=2 x 5!= 2 x (5 x 4 x 3 x 2 x 1)
\n=2 x 120
\n=240<\/p>\n

Question 27.
\nIn how many of the distinct permutations of the letters in MISSISSIPPI do the four I\u2019s not come together?
\nAnswer :
\nGiven work is MISSISSIPPI, which has 11 letters of which M-1, S-4, I-4, P-2.
\n\u2234 Total number of arrangements
\n\\(\\begin{aligned} &=\\frac{11 !}{4 ! 4 ! 2 !}\\&=\\frac{11 \\times 10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 !}{4 ! \\times(4 \\times 3 \\times 2 \\times 1) \\times(2 \\times 1)} \\end{aligned}\\)
\nTo find : Number of permutations of the letters in which four I’s not come together.
\nFirst, we find number of permutations of the letters in which four I’s are kept together.
\n\"1st
\nNow, we treat four P s together as single object this single object together with 7 remaining objects (letters) will account for 8 objects. These 8 objects in which there are four 5\u2019s and two P\u2019s can be arranged in \\(\\frac{8 !}{4 ! 2 !}\\) ways.
\n\u2234 Number of words in which I’s kept together.
\n\"1st<\/p>\n

Question 28.
\nIn how many ways can the letters of the word PERMUTATIONS be arranged if,
\n(i) words start with P and end with 5,
\n(ii) vowels are all together
\n(iii) There are always 4 letters between P and 5
\nAns:
\nIn the word PERMUTATION, there are 12 letters of which T-2
\n\"1st
\nWhen P and S are fixed in first and last places, we have to arrange remaining
\nletters in 10 places can be done in \\(\\frac{10 !}{2 !}\\) ways
\n\u2234 Required number of words
\n\\(\\frac{10 !}{2 !}\\)
\n\\(=\\frac{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 !}{2 !}\\)
\n= 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3
\n= 1814400<\/p>\n

\"KSEEB<\/p>\n

(ii) Vowels in the given word are A, E, I, O, U (5 letters) and remaining letters are seven in which T’s are two.
\nSince vowels are all together, so we treat them as a single object this single object together with 7 remaining objects (letters) will account for 8 objects. These 8 objects can be arranged in \\(\\frac{8!}{2 !}\\) ways.
\n\"1st
\nCorresponding to each of these arrangements, 5 vowels can be rearranged in. 5! Ways
\n(5<\/sup>P5<\/sub> ways)
\n\u2234 Required number of arrangements
\n\"1st<\/p>\n

(iii) Here, we want always 4 letters between P and S
\n\u2234 P and S can occupy 1st<\/sup> and 6th<\/sup> places
\nor 2nd<\/sup> and 7th<\/sup> places
\nor 3rd<\/sup> and 8th<\/sup> places
\nor 4th<\/sup> and 9th<\/sup> places
\nor 5th<\/sup> and 10th<\/sup> places
\nor 6th<\/sup> and 11th<\/sup> places
\nor 7th<\/sup> and 12th<\/sup> places
\n\u2234 P and S can be placed in 7 ways
\nSimilarly S and P can be placed in 7 ways. The remaining 10 places can be filled with remaining 10 letters in \\(\\frac{8!}{2 !}\\)ways (\u2235 T-2)
\n\u2234 Required arrangement
\n=7x2x\\(\\frac{8!}{2 !}\\)
\n=7×10!=25401600<\/p>\n

Question 29.
\nFind the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that
\n(i) all vowels occur together
\n(ii) all vowels do not occur together
\nAns:
\nGiven work DAUGHTER has 8 letters in which A, E,U are 3 vowels
\n\"1st<\/p>\n

We treat all vowels occur together as a single object and this single object together with 5 remaining objects (letters) will account for 6 object’s:’ These 6 objects can be arranged in 6<\/sup>P6<\/sub> ways = 6! Ways corresponding to each of these arrangements, 3 vowels can be rearranged in 3! Ways (3<\/sup> P3<\/sub> ways)
\n\u2234 Required number of arrangements
\n= (6 x 5 x 4 x 3 x 2 x 1) x (3 x 2 x l)
\n= 720 x 6 = 4320<\/p>\n

(ii) Required number of arrangements in which all vowels donot occur together = All possible arrangements of 8 letters taken all at time – number of arrangements in which the vowels occur together
\n= 8<\/sup>R8<\/sub>-6! x 3!
\n= 8! – 6! x 3!
\n= 61(8 x 7-3 x 2 x 1)
\n= 720 (50) = 36000<\/p>\n

Question 30.
\nIn how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?
\nAnswer :
\nTotal number of discs = 4 + 3 + 2 = 9
\nOut of nine, 4 are of one king, 3 are of second kind and 2 are of third kind
\n\u2234 Required arrangements
\n\\(=\\frac{9 !}{4 ! 3 ! 2 !}\\)
\n\\(=\\frac{9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 !}{4 ! \\times(3 \\times 2 \\times 1) \\times(2 \\times 1)}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 31.
\nFind the number of arrangements of the letters of the word In how many of these arrangement
\n(i) do the words start with P
\n(ii) do all the vowels always together
\n(iii) do the vowels never occur together
\n(iv) do the words begin with I and end in (Ex. No. 16. Text book)
\nAnswer:
\n\"1st<\/p>\n

Combination : Observations :<\/p>\n

(i) The formula for finding the number of combinations \u2018n\u2019 different objects taken V at a time is n<\/sup>Cr
\n<\/sub>(ii) In a combination order is not important
\n(iii) \\(^{n} P_{r}=^{n} C_{r} \\cdot r !, 0<r \\leq n\\)
\ni.e., total number of permutations of V different things (objects) taken V at a time is
\nn<\/sup>Cr <\/sub>r!= n<\/sup>Pr
\n<\/sub>\"1st<\/p>\n

Question 1.
\nShow that n<\/sup>Cr<\/sub> + n<\/sup>CT–<\/span><\/sub>x<\/sub> = n+1<\/sup>Cr<\/sub>.
\nAnswer :
\n\"1st<\/p>\n

Question 2.
\n\\(\\text { If }^{n} C_{8}=^{n} C_{2}, \\text { find }^{n} C_{2}\\)
\nAnswer:
\n\"1st<\/p>\n

Question 3.
\nDetermine
\n\\( ^{n} C_{17}, \\text { if }^{n} C_{9}=^{n} C_{8}\\)
\nAnswer:
\n\"1st<\/p>\n

Question 4.
\nDetermine n if
\n(i) \\( ^{2 n} C_{3}:^{n} C_{3}=12: 1\\)
\n(ii) \\( ^{2 n} C_{3}:^{n} C_{3}=11: 1\\)
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 5.
\nHow many chords can be drawn through 21 points on a circle?
\nAnswer :
\nWe know that, chord is the line segment joining any two points on the circumference of a circle.
\n\u2234 Number of chords is equal to the number of ways in which we can choose 2 points out of 21 points.
\n\u2234 Required number of chords
\n= 21<\/sup>C2
\n<\/sub>\\(=\\frac{21 \\times 20}{2 \\times 1}=210\\)<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIn how many ways can a team of 3 boys and 3 girls be selected form 5 boys and 4 girls?
\nAnswer :
\nWe have,
\nNumber of girls = 4
\nNumber of boys = 5
\n\u2234 3 girls out of 4 girls can be selected in 4<\/sup>C3 <\/sub>ways and 3 boys out of 5 boys can be selected in 5<\/sup>C3<\/sub> ways.
\n\u2234 Required number of ways
\n5<\/sup>C3 <\/sub>x 4<\/sup>C3<\/sub>
\n\\(=\\frac{5 \\times 4 \\times 3}{3 \\times 2 \\times 1} \\times \\frac{4 \\times 3 \\times 2}{3 \\times 2 \\times 1}\\)
\n=10 x 4 = 40<\/p>\n

Question 7.
\nFind the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each choice?
\nAnswer:
\nWe have
\nNumber of red balls = 6
\nNumber of White balls = 6
\nNumber of blue balls = 6
\nSince each selection consists of 3 balls of each = 11\u00a0 colours, so 3 red balls can be selected out of 6 balls in 6<\/sup>C3<\/sub> ways, 3 white balls out of 5 can be selected in\u00a05<\/sup>C3<\/sub> ways and 3 blue balls out of 5 can be selected in 5<\/sup>C3<\/sub> ways
\n\u2234 Required number of ways = 6<\/sup>C3<\/sub> x 5<\/sup>C3<\/sub> x 5<\/sup>C3
\n<\/sub>\\(=\\frac{6 \\times 5 \\times 4}{3 \\times 2 \\times 1} \\times \\frac{5 \\times 4 \\times 3}{3 \\times 2 \\times 1} \\times \\frac{5 \\times 4 \\times 3}{3 \\times 2 \\times 1}\\)
\n= 20 x 10 x 10=2000<\/p>\n

Question 8.
\nDetermine the number of 5 cards combinations out of deck of 52 cards if there is exactly one ace is each combination.
\nAnswer:
\nWe have,
\nNumber of cards in a deck = 52.
\nIf 52 cards, there are 4 aces and 48 other cards, (\u2235 4 + 48 = 52).
\nIf there is exactly one ace in each 5 card combination, then one ace out of 4 can be selected in 4<\/sup>C1 <\/sub>ways and 4 non-ace cards can be selected out of 48 in 48<\/sup> C4<\/sub> ways.
\n\u2234 Required number of combination = 4<\/sup> C1<\/sub> x 48<\/sup>C4
\n<\/sub>\\(=\\frac{4}{1} \\times \\frac{48 \\times 47 \\times 46 \\times 45}{4 \\times 3 \\times 2 \\times 1}=778320\\)<\/p>\n

Question 9.
\nWhat is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
\n(i) four cards are of the same suit
\n(ii) four cards belong to four different suit
\n(iii) are face cards
\n(iv) two red cards and two are black card
\n(v) 4 cards are of the same colour.
\nAnswer:
\nWe have,
\nNumber of ways of choosing 4 cards from a pack of 52 cards is the number of combinations of 52 different things taken 4 at time
\n\u2234 Required number of ways = 52<\/sup>C4
\n<\/sub>\\(=\\frac{52 \\times 51 \\times 50 \\times 49}{4 \\times 3 \\times 2 \\times 1}=270725\\)<\/p>\n

(i) There are four suits, namely diamond, club, spade, heart and each suit has 13 cards. We have to choose 4 cards of the same suit so 4 diamond cards out of 13 diamond cards can be selected in 13<\/sup>C4<\/sub> ways. Similarly,there are 13<\/sup> C4<\/sub> ways of choosing 4 spades and 13<\/sup>C4<\/sub> ways of choosing 4 hearts.
\n\u2234 Required number of ways
\n= 13<\/sup>c4<\/sub>+13<\/sup>c4<\/sub>+13<\/sup>c4<\/sub>+13<\/sup>c4
\n<\/sub>\\(=4 x^{13} C_{4}=4 \\times \\frac{13 \\times 12 \\times 11 \\times 10}{4 \\times 3 \\times 2 \\times 1}=2860\\)<\/p>\n

(ii) There are 13 cards in each suit we have to select 4 cards belonging to 4 different suits. There are l3<\/sup>C1<\/sub>\u00a0ways of choosing one card from 13 diamond cards, 13<\/sup>C1<\/sub>\u00a0ways of choosing 1 card from 13 cards of spades, 13<\/sup>C1<\/sub> ways of choosing 1 card from 13 cards of club and 13<\/sup>C1<\/sub>\u00a0ways of choosing 1 card from 13 cards of hearts.
\nBy multiplication principle, the required number of ways
\n= 13<\/sup>C1 <\/sub>x 13<\/sup>C1<\/sub> x 3<\/sup>C1<\/sub>x 3<\/sup>C1
\n<\/sub>= 13 x 13 x 13 x 13 = 134<\/sup> =28561<\/p>\n

\"KSEEB<\/p>\n

(iii) Face cards means Kings, Queen, Jack there are 4 suits, each suit has 3 face cards. Therefore there are 12 face cards and 4 are to be selected out of 12 cards, in 12<\/sup>C4<\/sub> ways
\nRequired number of ways = 12<\/sup>C4
\n<\/sub>\\(=\\frac{12 \\times 11 \\times 10 \\times 9}{4 \\times 3 \\times 2 \\times 1}=495\\)<\/p>\n

(iv) There are 26 red cards and 26 black cards.
\n\u2234 We have select 2 red and 2 black cards. This can be done in
\n26<\/sup>C2<\/sub> x 26<\/sup>C2 <\/sub>ways.
\n\\(\\begin{aligned} &=\\frac{26 \\times 25}{2 \\times 1} \\times \\frac{26 \\times 25}{2 \\times 1}\\\\ &=325 \\times 325=105625 \\end{aligned}\\)
\n\u2234 Required number of ways = 26<\/sup>C2 <\/sub>x 26<\/sup>C2
\n<\/sub>\\(\\begin{aligned}&=\\frac{26 \\times 25}{2 \\times 1} \\times \\frac{26 \\times 25}{2 \\times 1}\\\\&=325 \\times 325=105625\\end{aligned}\\)<\/p>\n

(v) There are 26 red and 26 black cards
\n\u2234 4 red cards can be selected in 26<\/sup> C4<\/sub> ways 4 black cards can be selected in 26<\/sup> C4<\/sub> ways
\n\u2234 Required number of ways = 26<\/sup> C4<\/sub> +26<\/sup>C4
\n<\/sub>\\(=2 x^{26} C_{4}=\\frac{26 \\times 25 \\times 24 \\times 26}{4 \\times 3 \\times 2 \\times 1}=29900\\)<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIn how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers?
\nAnswer:
\nWe have,
\nNumber of players = 17
\nNumber of players to be select for team =11
\nNumber of bowlers = 5
\nNumber of bowlers to be select = 4
\nNumber of players other than bowlers =12
\n4 bowlers out of 5 can be selected in 5<\/sup>C4<\/sub> ways and 7(i.e., 11-4 = 7) other players can be selected out of 12 (i.e., 17 – 5 = 12) in 12<\/sup>C7<\/sub> ways.
\n\u2234 Required number of ways = 5<\/sup>C4 <\/sub>x 12<\/sup>C7
\n<\/sub>\\(=^{5} C_{1} \\times^{12} C_{5} \\quad ^{n} C_{r}=^{n} C_{n-r}\\)
\n\\( =5 \\frac{12 \\times 11 \\times 10 \\times 9 \\times 8}{5 \\times 4 \\times 3 \\times 2 \\times 1}=3960\\)<\/p>\n

Question 11.
\nA bag contains 5 black and 6 red balls. Determine the number of ways which 2 black and 3 red balls can be selected.
\nAnswer:
\n2 black balls out of 5 can be selected in 5<\/sup>C2 <\/sub>ways and 3 red balls out of 6 can be selected in 6<\/sup> C3 <\/sub>ways.
\n\u2234 Required number of ways = 5<\/sup>C2<\/sub> x 6<\/sup>C3
\n<\/sub>\\(=\\frac{5 \\times 4}{2 \\times 1} \\times \\frac{6 \\times 5 \\times 4}{3 \\times 2 \\times 1}=200 \\)<\/p>\n

Question 12.
\nIn how many ways can be student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
\nAnswer:
\nWe have,
\nNumber of available courses = 9
\nNumber of compulsory courses = 2
\n\u2234 Number of remaining courses available for choice
\n=9-2=7
\nNumber of courses to be choose = 5
\nBut 2 courses are compulsory, so the student can select
\n3 courses out of remaining of courses in 7<\/sup>C3<\/sub> ways.
\nRequired number of ways = 7<\/sup>C3
\n<\/sub>\\(=\\frac{7 \\times 6 \\times 5}{3 \\times 2 \\times 1}=35\\)<\/p>\n

Question 13.
\nA committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees should consists of 1 man and 2 women?
\nAnswer:
\nWe have,
\nNumber of men = 2
\nNumber of women = 3
\n\u2234 Total number of persons = 2 + 3 = 5
\nHere, number of persons to be select = 3<\/p>\n

(i) Required number of ways of selecting 3 persons
\n\\(^{5} C_{3}=\\frac{4 \\times 5}{2}=10\\)<\/p>\n

Alternatives:
\n\"1st<\/p>\n

(ii) 1 man can select out of 2 men in 2<\/sup>C, ways and 2 women can select out of 3 women in 3<\/sup>C2<\/sub> ways.
\n\u2234 Required number of ways = 2<\/sup>C1 <\/sub>x 3<\/sup>C2<\/sub> = 2 x 3 = 6<\/p>\n

Miscellaneous Problems<\/strong><\/p>\n

Question 1.
\nHow many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
\nAnswer:
\nGiven word is DAUGHTER Here, vowels are A, E, U (3 in number)
\nConsonant are D, G, H, T, R (5 in number)
\n\u2234 2 vowels out 3 and 3 consonant out of 5 can be selected in 3<\/sup>C2 <\/sub>x 5<\/sup>C3<\/sub> ways and each
\nselection of 2 vowels and 3 consonants can be arranged in 5<\/sup>P5<\/sub> =5! ways.
\nRequired number of words = 3<\/sup>C2 <\/sub>x 5<\/sup>C3 <\/sub>x 5!
\n\\(=3 \\times \\frac{5 \\times 4}{2 \\times 1} \\times 120=3600\\)<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nHow many words, with or without meaning can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together.
\nAnswer:
\nGiven word is EQUATION
\nIn this word, there are five voweIs A, E, I, O, U and three consonants Q, T, N
\n\"1st<\/p>\n

Question 3.
\nA committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of
\n(i) Exactly 3 girls ?
\n(ii) atleast 3 girls ?
\n(iii)atmost 3 girls
\nAnswer:
\nNumber of girls = 4
\nNumber of boys = 9
\n(i) Since the committee has to consists exactly 3 girls, the committee can consist of 3 girls and 4 boys.
\n\u2234 3 girls out of 4 can be selected in 4<\/sup>C3<\/sub> ways out of 4 boys out of 9 can be
\nselected in 9<\/sup>C4 <\/sub>ways
\n\u2234 Required number of ways = 4<\/sup>C3 <\/sub>x 9<\/sup>C4
\n<\/sub>\\(=4 \\times \\frac{9 \\times 8 \\times 7 \\times 6}{4 \\times 3 \\times 2 \\times 1}=504\\)<\/p>\n

(ii) Since the committee has to consist at least 3 girls, the committee can consists of
\n(a) 3 girls and 4 boys
\n(b) 4 girls and 3 boys
\n(a) 3 girls and 4 boys can be selected in = 4<\/sup>C3 <\/sub>x 9<\/sup>C4 <\/sub>ways
\n(b) 4 girls and 3 boys can be selected in 4<\/sup>C4 <\/sub>x 9<\/sup>C3 <\/sub>ways
\n\u2234 Required number of ways
\n= 4<\/sup>C3 <\/sub>x 9<\/sup>C4 <\/sub>+ 4<\/sup>C4 <\/sub>x 9<\/sup>C3<\/sub>
\n4 x 126 + 1 x 84 = 504 + 84 = 588<\/p>\n

(iii) In this case, commitee may consists of 3 girls, 2 girls ,1 girl or no girl.
\n\"1st<\/p>\n

Question 4.
\nIf the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
\nAnswer:
\nWe are required to find the number of permutations which begin with the letter A.
\nThere are 11 letters in the given word of which A – 2, 1-2,N-2,
\nWhen A is fixed in the first place, remaining 10 \\(\\frac{10 !}{2 ! 2 !}\\)letters can be arranged in ways.
\n\u2234 Required permulations
\n\\(=\\frac{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3}{2 \\times 1}=907200\\)<\/p>\n

Question 5.
\nHow many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
\nAnswer:
\nGiven digits are 0, 1, 3, 5, 7, 9
\nRequired number is 6 – digit which is divisible by 10.
\n\"1st
\nWe know that, the number is divisible 10, the unit\u2019s place should be compulsorily filled by 0. In 1<\/sup>P1<\/sub> ways. The remaining 5 places can be filled in 5<\/sup>P5<\/sub> ways.
\n= 5<\/sup>P5<\/sub> x 1<\/sup>P1 <\/sub>=120 x 1= 120.<\/p>\n

Question 6.
\nThe English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
\nAnswer:
\n\"1st
\nGiven: 5 vowels, 21 consonants
\nRequired: Four letter words with 2 different vowels and 2 different consonants
\n\u2234 The number of ways of selecting 2 vowels out of 5= 5<\/sup>C2<\/sub>=10
\nThe number of ways of selecting 2 consonants out of 21 =21<\/sup>C2<\/sub> = 210
\nNumber of combinations of 2 vowels and 2 consonants is 10 x 210 = 2100.
\nNow, each of these 2100 combinations has 4 letters which can be arranged themselves in 4<\/sup>P4<\/sub> = 4! ways.
\nRequired number of words = 2100 x 4!
\n= 2100 x 24 = 50400.<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nIn an examination, a question paper consists of 12 questions divided into two parts, part I and part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting atleast 3 from each part. In how many ways can a student select the questions?
\nAnswer:
\nWe have,
\nNumber of questions in the question paper = 12
\nNumber of questions in part 1 = 5
\nNumber of question in part II = 7
\nNumber of questions to attempt = 8
\nCondition: Selecting at least 3 from each part<\/p>\n\n\n\n\n\n\n
Part I (out of 5)<\/td>\nPart II (out of 7)<\/td>\nNumber of ways<\/td>\n<\/tr>\n
3<\/td>\n5<\/td>\n5<\/sup>C3<\/sub> x 7<\/sup>C5<\/sub><\/td>\n<\/tr>\n
4<\/td>\n4<\/td>\n5<\/sup>C4 <\/sub>x 7<\/sup>C4<\/sub><\/td>\n<\/tr>\n
5<\/td>\n3<\/td>\n5<\/sup>C5<\/sub> x 7<\/sup>C3<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"1st<\/p>\n

Question 8.
\nDetermine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king?
\nAnswer:
\nWe have to select one king out of 4 kings and four other cards out of 48. This can be done in 4<\/sup>C,x48<\/sup>C4<\/sub> ways
\n\\(=\\frac{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3}{2 \\times 1}=907200\\)
\n\u2234 Required number of ways = 4<\/sup>C1<\/sub> x 48<\/sup>C4
\n<\/sub>=4 x 2 x 47 x 46 x 45 = 778320<\/p>\n

Question 9.
\nIt is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
\nAnswer:
\nWe have,
\nNumber of seats = 9
\n\"1st
\nHere, 4 even places and 5 odd places.
\n\u2234 4 even places can be filled with 4 women in 4<\/sup>P4<\/sub> = 4! ways and 5 odd places can be filled with 5 men in 5<\/sup>P5<\/sub> =5! ways. Required number of ways
\n= 5! x 4!
\n= 120 x 24 = 2880<\/p>\n

Question 10.
\nIn how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
\nAnswer:
\nFirst, we seat the 5 girls in 5! ways. For each such a arrangement, the three boys can be seated only at the \u20186\u2019 cross marks places x G x G x G x G x G x Three boys can be selected in 6<\/sup> P3<\/sub> ways Hence total number of ways
\n= 5 x 6<\/sup>P3 <\/sub>= (5 x 4 x 3 x 2 x 1) x (6 x 5 x 4)
\n= 14400<\/p>\n

Question 11.
\nIn how many ways can be letters of the word ASSASSINATION be arranged so that all the S\u2019s are together?
\nAnswer:
\nIn the given word there are 13 letters of which S-4, A-3,1-2, N-2,0-1, T-l.
\nSince all four S\u2019s are together then we treat them as a single object (letter). This single object (letter)
\n\u2234 These 10 items can be arranged in \\(\\frac{10 !}{2 ! 2 !}\\)
\n\u2234 Required numbers of words
\n\\(=\\frac{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3}{2 \\times 1}=907200\\)<\/p>\n

Question 12.
\nIn how many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
\nAnswer:
\nIn the word INVOLUTE there are 4 vowels (I, O, E, U) and 4 consonants (N, V, L, T)
\nNumber of ways of selecting 3 vowels out of 4 = 4<\/sup>C3<\/sub> = 4
\nNumber of ways of selecting 2 consonants out of 4 = 4<\/sup>C2<\/sub> =6
\n\u2234 Number of combinations of 3 vowels and 2 consonants is 4 x 6 = 24
\nNow, each of these 24 combinations has 5 letters Which can be arranged themselves in 5! Ways
\n\u2234 Required number words = 24 x 5 !
\n= 24 x 120 = 2880.<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nA group consists of 4 girls and 7 boys. In how many ways can a team of 5 members selected if the team has
\n(i) no girl
\n(ii) at least 3 girls
\n(iii) at least one boy and one girl
\nAnswer:
\n(i) no girl
\nTable for computing the number of ways of selections.<\/p>\n\n\n\n\n
Girls (out of 4)<\/td>\nBoys (out of 7)<\/td>\nNumber of ways<\/td>\n<\/tr>\n
0<\/td>\n5<\/td>\n4<\/sup>C0 <\/sub>x 7<\/sup>C5<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Required number of ways = 4<\/sup>C0 <\/sub>x 7<\/sup>C5
\n<\/sub>\\(=1 \\times^{7} C_{2}=\\frac{7 \\times 6}{2 \\times 1}=21\\)<\/p>\n

(ii) at least 3 girls<\/p>\n\n\n\n\n\n
Girls (out of 4)<\/td>\nBoys (out of 7)<\/td>\nNumber of ways<\/td>\n<\/tr>\n
3<\/td>\n2<\/td>\n4<\/sup>C3 <\/sub>x 7<\/sup>C2<\/sub><\/td>\n<\/tr>\n
4<\/td>\n1<\/td>\n4<\/sup>C4 <\/sub>x 7<\/sup>C1<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"1st<\/p>\n

(iii) at least one boy and one girl<\/p>\n\n\n\n\n\n\n\n
Girls (out of 4)<\/td>\nBoys (out of 7)<\/td>\nNumber of ways<\/td>\n<\/tr>\n
1<\/td>\n4<\/td>\n4<\/sup>C1<\/sub> X 7<\/sup>C4<\/sub> =140<\/td>\n<\/tr>\n
2<\/td>\n3<\/td>\n4<\/sup>C2 <\/sub>x 7<\/sup>C3<\/sub> = 210<\/td>\n<\/tr>\n
3<\/td>\n2<\/td>\n4<\/sup>C3 <\/sub>x 7<\/sup>C2<\/sub> =84<\/td>\n<\/tr>\n
4<\/td>\n1<\/td>\n4<\/sup>C4 <\/sub>x 7<\/sup>C1<\/sub> = 07<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Required numbers of ways
\n= 140 + 210+84 + 07 = 441<\/p>\n

Question 14.
\nFind the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary what will be the 50th <\/sup>word?
\nAnswer:
\nThere are 5 letters in the word AGAIN, in which A – 2
\n\u2234 Required numbers of ways
\n\\(=\\frac{5 !}{2 !}=\\frac{5 \\times 4 \\times 3 \\times 2 \\times 1}{2 \\times 1}=60\\)
\nTo get the number of words starting with A, fix A in first place and remaining 4 letters can be arranged in 4! Ways = 24 ways
\nTo get the number of word starting with G, fix G in first place and the remaining 4 letters in which A – 2
\n\u2234 Number of words starting with \\(G=\\frac{4 !}{2 !}=12 \\)
\nSimilarly ,numbers of words starting with \\(I=\\frac{4 !}{2 !}=12\\)
\nTotal number of words so far obtained = 24 + 12 + 12 = 48
\nThe 49th<\/sup> word is NAAGI and 50th<\/sup> word is NAAIG<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nHow many numbers greater than 1000000 can be formed by using the digits
\n1, 2,0,2,4,2,4?
\nAnswer:
\nSince the required numbers are greater than 1000000, then the number is begin with either 1,2 or 4
\n\u2234 Number of numbers begin with 1
\n\\(=\\frac{6 !}{3 ! 2 !}=\\frac{6 \\times 5 \\times 4 \\times 3 !}{3 !(2 \\times 1)}=60\\)<\/p>\n

Total numbers beginning with 2
\n\\(=\\frac{6 !}{2 ! 2 !}=\\frac{3 \\times 4 \\times 5 \\times 6}{2}=180\\)<\/p>\n

Total numbers beginning with 4
\n\\(=\\frac{6 !}{3 !}=6 \\times 5 \\times 4=120\\)<\/p>\n

Required number of numbers
\n= 60+ 180 + 120 = 360<\/p>\n

Alternative
\nRequired number of numbers
\n\\(=\\frac{7 !}{3 ! 2 !}-\\frac{6 !}{3 ! 2 !}=420-60=360\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 7 Permutations and Combinations Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Maths Question Bank Chapter 7 Permutations and Combinations Question 1. State fundamental principle of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13940"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=13940"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13940\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}