Question

The number of selection of $n$ objects from $2n$ objects of which $n$ are identical and the rest are different is

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

Answer 1

Answer: (A)

$2^n$

Let $S_1$ be the set of $n$ identical objects and $S_2$ be the set of the remaining $n$ different objects.

Number of ways to select $r$ objects from $S_1=1$ since all objects in $S_1$ are identical.

Number of ways to select $r$ objects from $S_2{=}^n{C}_r$.

Total number of ways to select $n$ objects =

$\bullet$$n$ objects from $S_2$ and $0$ object from $S_1$

Number of ways ${=}^n{C}_n$

$\bullet$$n-1$ objects from $S_2$ and $1$ object from $S_1$

Number of ways ${=}^n{C}_{n-1}\times 1{=}^n{C}_{n-1}$

$\bullet$ $n-2$ objects from $S_2$ and $2$ object from $S_1$

Number of ways ${=}^n{C}_{n-2}\times 1{=}^n{C}_{n-2}$

$\bullet$ $n-3$ objects from $S_2$ and $3$ object from $S_1$

Number of ways ${=}^n{C}_{n-3}\times 1{=}^n{C}_{n-3}$

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$\bullet$ $2$ objects from $S_2$ and $n-2$ object from $S_1$

Number of ways ${=}^n{C}_2\times 1{=}^n{C}_2$

$\bullet$ $1$ objects from $S_2$ and $n-1$ object from $S_1$

Number of ways ${=}^n{C}_1\times 1{=}^n{C}_1$

$\bullet$ $0$ objects from $S_1$ and $n$ object from $S_1$

Number of ways ${=}^n{C}_0\times 1{=}^n{C}_0$

$\therefore$ Total number of ways ${=}^n{C}_n{+}^n{C}_{n-1}{+}^n{C}_{n-2}{+}^n{C}_{n-3}+....{+}^n{C}_2{+}^n{C}_1{+}^n{C}_0=2^n$

So, the answer is option (A).