The number of ways of choosing 10 objects out | vedclass

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(A) Let the $10$ identical objects be $I$ and the $21$ distinct objects be $D_1, D_2, ..., D_{21}$.We need to choose $10$ objects in total.Let $k$ be

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$2^{20}$

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(A) Let the $10$ identical objects be $I$ and the $21$ distinct objects be $D_1, D_2, ..., D_{21}$.We need to choose $10$ objects in total.Let $k$ be the number of distinct objects chosen,where $0 \le k \le 10$.The number of ways to choose $k$ distinct objects from $21$ is $^{21}C_k$.Once $k$ distinct objects are chosen,the remaining $(10-k)$ objects must be chosen from the $10$ identical objects. Since the objects are identical,there is only $1$ way to choose them.Thus,the total number of ways is the sum of $^{21}C_k$ for $k=0$ to $10$:Total ways $= \sum_{k=0}^{10} {^{21}C_k}$.We know that $\sum_{k=0}^{21} {^{21}C_k} = 2^{21}$.Also,by symmetry,$\sum_{k=0}^{10} {^{21}C_k} = \sum_{k=11}^{21} {^{21}C_k}$.Let $S = \sum_{k=0}^{10} {^{21}C_k}$. Then $2S = \sum_{k=0}^{10} {^{21}C_k} + \sum_{k=11}^{21} {^{21}C_k} = \sum_{k=0}^{21} {^{21}C_k} = 2^{21}$.Therefore,$S = \frac{2^{21}}{2} = 2^{20}$.

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

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