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Question

The maximum number of permutations of 2n letters in which there are only as and bs, taken all at a time is given by

A
2nCn
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B
21621034n6n14n2n
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C
n+11n+22n+33n+442n1n12nn
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D
2n[135(2n3)(2n1)]n!
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Solution

The correct option is D 2n[135(2n3)(2n1)]n!
Let the number of as be x.
Then, number of bs is 2nx
As as and bs are alike, number of permutations =(2n)!x! (2nx)!=2nCx

Now, we know that, if n is even, nCr is maximum for r=n2
2nCx is maximum when r=2n2=n
Hence, maximum number of permutations =2nCn
2nCn=(2n)!n! n!
=2n(2n1)(2n2)(n+1)n!n! n!
=2n(2n1)(2n2)(n+1)n!
=(n+1)(n+2)(n+3)(2n1)(2n)123(n1)n

2nCn=2n!n!n!

=123(2n1)(2n)n!n!

=(135(2n1))(2462n)n!n!

=135(2n1)2n(123n)n!n!
=2n(135(2n3)(2n1)n!
=2610(4n2)123(n)
(multiplying whole numerator by 2n)
=21621034n6n14n2n

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