Question

The number of ways of choosing $n$ objects out of $(3n+1)$ objects of which $n$ are identical and $(2n+1)$ are distinct, is

• A. $2^{2n}$
• B. $2^{2n+1}$
• C. $2^{2n-1}$
• D. none of these

Answer 1

The correct option is A $2^{2n}$

If we choose k $(0\le k\le n)$ identical objects, then we must choose $(n-k)$ distinct objects. This can be done in ${{}^{2n+1}C}_{n-k}$ways. Thus the required number of ways

$={\sum }_{k=0}^n{{}^{2n+1}C}_{n-k}$

$={{}^{2n+1}C}_n+{{}^{2n+1}C}_{n-1}+{{}^{2n+1}C}_{n-2}....+{{}^{2n+1}C}_0$

$=2^{2n}$