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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a
4n!(n!)2
b
4n!2n!
c
4n!n!
d
3n!(n!)2
answer is D.
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Detailed Solution
Cn 3n×(2n)!n!=3n!(n!)2 (Selection and Arrangement)