The number of ways of choosing 10 objects out | vedclass

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(A) Let the $10$ identical objects be denoted by $I$ and the $21$ distinct objects be $D_1, D_2, \dots, D_{21}$.To choose $10$ objects,we can choose $

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$2^{20}$

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(A) Let the $10$ identical objects be denoted by $I$ and the $21$ distinct objects be $D_1, D_2, \dots, D_{21}$.To choose $10$ objects,we can choose $k$ distinct objects from the $21$ distinct objects,where $0 \le k \le 10$.The remaining $(10-k)$ objects must be chosen from the $10$ identical objects.Since the $10$ identical objects are indistinguishable,there is only $1$ way to choose any number of them.Thus,the total number of ways is the sum of the ways to choose $k$ distinct objects for all possible values of $k$ from $0$ to $10$:Total ways $= \sum_{k=0}^{10} {}^{21}C_k$.We know that $\sum_{k=0}^{21} {}^{21}C_k = 2^{21}$.Since ${}^{21}C_k = {}^{21}C_{21-k}$,we have $\sum_{k=0}^{10} {}^{21}C_k = \sum_{k=11}^{21} {}^{21}C_k$.Let $S = \sum_{k=0}^{10} {}^{21}C_k$. Then $S + S = \sum_{k=0}^{21} {}^{21}C_k = 2^{21}$.Therefore,$2S = 2^{21}$,which implies $S = 2^{20}$.

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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