There are 4n things of which n are alike and all the rest different. Then the number of permutations of 4n things taken 2n at a time, each permutation containing the n like things

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4n!(n!)2

4n!2n!

4n!n!

3n!(n!)2

answer is D.

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Detailed Solution

Cn 3n×(2n)!n!=3n!(n!)2 (Selection and Arrangement)

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