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 $D_1, D_2, \dots, D_{21}$.To choose $10$ objects,we can choose $k$ objects

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$2^{20}$

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(A) Let the $10$ identical objects be $I$ and the $21$ distinct objects be $D_1, D_2, \dots, D_{21}$.To choose $10$ objects,we can choose $k$ objects from the $21$ distinct ones,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 $\sum_{k=0}^{10} \binom{21}{k}$.We know that $\sum_{k=0}^{21} \binom{21}{k} = 2^{21}$.Since $\binom{21}{k} = \binom{21}{21-k}$,we have $\sum_{k=0}^{10} \binom{21}{k} = \sum_{k=11}^{21} \binom{21}{k}$.Let $S = \sum_{k=0}^{10} \binom{21}{k}$. Then $2S = \sum_{k=0}^{10} \binom{21}{k} + \sum_{k=11}^{21} \binom{21}{k} = \sum_{k=0}^{21} \binom{21}{k} = 2^{21}$.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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