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Question: The number of ways of selecting n objects out of 3n objects, of which n are alike and rest are diffe...

The number of ways of selecting n objects out of 3n objects, of which n are alike and rest are different is –

A

22n–1 +(2n−1)!n!(n−1)!\frac{(2n - 1)!}{n!(n - 1)!}

B

22n–1 – (2n−1)!n!(n−1)!\frac{(2n - 1)!}{n!(n - 1)!}

C

22n+1 + (2n+1)!n!(n+1)!\frac{(2n + 1)!}{n!(n + 1)!}

D

None of these

Answer

22n–1 +(2n−1)!n!(n−1)!\frac{(2n - 1)!}{n!(n - 1)!}

Explanation

Solution

The required number of ways

= Coefficient of xn in (x0 + x1 + x2 + … + xn) (x0 + x)2n

= Coefficient of xn in (1−xn+11−x)\left( \frac{1 - x^{n + 1}}{1 - x} \right) (1 + x)2n

= Coefficient of xn in (1 – x)–1 (1 + x)2n

= Coefficient of xn in (∑r=0∞xr)\left( \sum_{r = 0}^{\infty}x^{r} \right) (1 + x)2n

= Coefficient of xn in (∑r=0nxr)\left( \sum_{r = 0}^{n}x^{r} \right) (1 + x)2n

= Coefficient of xn in ∑r=0nxr\sum_{r = 0}^{n}x^{r} (1 + x)2n

= ∑r=0n\sum_{r = 0}^{n}{}Coefficient of xn–r in (1 + x)2n

= ∑r=0n2nCn−r\sum_{r = 0}^{n}{2nC_{n - r}}

= 2nCn + 2nCn+1 + …. + 2nC1 + 2nC0

=12\frac { 1 } { 2 } [2nCn+{2nC0+2nC1+....+2nCn+2n+1Cn+1+....+2nC2n}]\left\lbrack 2nC_{n} + \left\{ 2nC_{0} +^{2n}C_{1} + .... +^{2n}C_{n} +^{2n + 1}C_{n + 1} + .... +^{2n}C_{2n} \right\} \right\rbrack

[Q nCr = nCn–r]

= 12\frac{1}{2} [2nCn + 22n]

= 12\frac { 1 } { 2 } [2n!n!n!+22n]\left\lbrack \frac{2n!}{n!n!} + 2^{2n} \right\rbrack

= 22n – 1 + (2n−1)!n!(n−1)!\frac{(2n - 1)!}{n!(n - 1)!}

Hence (1) is correct answer.