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Let N denotes the number of ways in whic...

Let N denotes the number of ways in which 3n letters can be selected from 2n A's, 2nB's and 2nC's. then,

A

`3|(N-1)`

B

`n|(N_1)`

C

`(n+1(|(N-1)`

D

`3n(n+1)|(N-1)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`becausex+y+z=3n`
`impliesN=`Coefficient of `alpha^(3n)` in `(1+alpha+alpha^(2)+ . . .+alpha^(2n))^(3)`
=Coefficient of `alpha^(3n)` in `(1-alpha^(2n+1))^(3)(1-alpha)^(-3)`
=Coeffieint of `alpha^(3n)` in `(1-3alpha^(2n-1))(1+.^(3)C_(1)alpha+ . . .)`
`=.^(3n+2)C_(3n-3)*.^(n+1)C_(n-1)`
`=.^(3n+2)C_(2)-3*.^(n+1)C_(2)=3n^(2)+3n+1`
`thereforeN-1=3n(n+1)`
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