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Permutations and Combinations question

2007 · Shift 0 · Q58
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Permutations and Combinations question

2007 · Shift 0 · Q58

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus A∪B∪C=S, A∩B=B∩C=A∩C=ϕA \cup B \cup C = S,\,A \cap B = B \cap C = A \cap C = \phiA∪B∪C=S,A∩B=B∩C=A∩C=ϕ. The number of ways to partition S is
  1. A
    12!(4!)3  {{12!} \over {{{(4!)}^3}}}\,\,(4!)312!​
  2. B
    12!(4!)4  {{12!} \over {{{(4!)}^4}}}\,\,(4!)412!​
  3. C
    12!3!  (4!)3{{12!} \over {3!\,\,{{(4!)}^3}}}3!(4!)312!​
  4. D
    12!3!  (4!)4{{12!} \over {3!\,\,{{(4!)}^4}}}3!(4!)412!​
View written solutionFree

Correct answer: A

  1. Since S=\{1,2,3,\dots,12\} is partitioned into three sets A,B,CA,B,CA,B,C of equal size, each set must contain 123=4\frac{12}{3}=4312​=4 elements.

  2. We need the number of ways to divide 12 distinct elements into 3 disjoint groups of 4 each.

  3. First, choose 4 elements for AAA: (124)\binom{12}{4}(412​)

  4. Then choose 4 elements for BBB from the remaining 8: (84)\binom{8}{4}(48​)

  5. The remaining 4 automatically go to CCC: (44)=1\binom{4}{4}=1(44​)=1

So total ways are (124)(84)(44)\binom{12}{4}\binom{8}{4}\binom{4}{4}(412​)(48​)(44​)

  1. Simplify: (124)(84)=12!4!8!⋅8!4!4!=12!(4!)3\binom{12}{4}\binom{8}{4} = \frac{12!}{4!8!}\cdot \frac{8!}{4!4!} = \frac{12!}{(4!)^3}(412​)(48​)=4!8!12!​⋅4!4!8!​=(4!)312!​

Thus, N=12!(4!)3N=\frac{12!}{(4!)^3}N=(4!)312!​

  1. Important note: We do not divide by 3!3!3! because the sets are specifically named A,B,CA,B,CA,B,C. Interchanging AAA and BBB gives a different assignment of elements to the labeled sets.

  2. Therefore the correct option is 12!(4!)3\boxed{\frac{12!}{(4!)^3}}(4!)312!​​ which is Option A.

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