Question

The number of ways of choosing $10$ objects out of $31$ objects of which $10$ are identical and the remaining $21$ are distinct, is

• A. $2^{20}-1$
• B. $2^{20}$
• C. $2^{20}+1$
• D. $2^{21}$

Answer 1

The correct option is B $2^{20}$

Given that, out of $31$ objects, $10$ are identical and remaining $21$ objects are distinct. So in following ways, we can choose $10$ objects out of $31$ objects.

$0$ identical $+10$ distincts, number of ways $=1\times {}^{21}{C}_{10}$
$1$ identical $+9$ distincts, number of ways $=1\times {}^{21}{C}_9$
$2$ identical $+8$ distincts, number of ways $=1\times {}^{21}{C}_8$
........ and so on ........
So, total number of ways in which we can choose $10$ objects is
${}^{21}{C}_{10}+{}^{21}{C}_9+{}^{21}{C}_8+\cdots +{}^{21}{C}_0=x\left(\text{Let}\right)\cdots \left(I\right)$
$\because {}^n{C}_r={}^n{C}_{n-r}$
${}^{21}{C}_{11}+{}^{21}{C}_{12}+{}^{21}{C}_{13}+\cdots +{}^{21}{C}_{21}=x\cdots \left(II\right)$
On adding both Equation $\left(I\right)$ and $\left(II\right)$, we get
$2x={}^{21}{C}_0+{}^{21}{C}_1+{}^{21}{C}_2+\cdots +{}^{21}{C}_{21}$
$\therefore 2x=2^{21}$
$\Rightarrow x=2^{20}$