Question

The numb er of ways of selecting $n$ things out of $3n$ things, of which $n$ are of one kind. and $n$ are of a second kind and the rest unlike,

• A. $n\cdot 2^{n-1}$
• B. $(n-1)\cdot 2^{n-1}$
• C. $(n+2)2^{n-1}$
• D. $(n+1)2^{n-1}$

Answer 1

The correct option is C $(n+2)2^{n-1}$

Suppose k objects have been chosen out of the n dissimilar objects.

Then we have to choose $n-k$ objects from out of the two other sets. The choices would go like O from the fist set and $(n-k)$ from other.

$\therefore \sum _{k=0}^n(n-k+1)\left(\frac{n}{k}\right)=\sum _{k=1}^n(n+1)\left(\begin{array}{c}n\\ k\end{array}\right)-\sum _{k=0}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)$

The first sum is just $(n+1)2^n$. The second sum

$\sum _{k=0}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)=\sum _{k=0}^n(n-k)\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{1}{2}\sum _{k=0}^{n}n\left(\begin{array}{c}n\\ k\end{array}\right)=n\times 2^{n-1}$

Hence the required number of ways are $(n+2)2^{n-1}$.