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

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$2^{20}$

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(A) Let the $10$ identical objects be $I$ and the $21$ distinct objects be $D_1, D_2, ..., D_{21}$.To choose $10$ objects,we can select $k$ objects from the $21$ distinct objects and $(10-k)$ objects from the $10$ identical objects,where $0 \le k \le 10$.Since the $10$ identical objects are indistinguishable,there is only $1$ way to choose any number of them.Thus,the number of ways is the sum of the ways to choose $k$ distinct objects for all possible $k$ from $0$ to $10$:Total ways = $\sum_{k=0}^{10} \binom{21}{k} = \binom{21}{0} + \binom{21}{1} + ... + \binom{21}{10}$.We know that the sum of binomial coefficients is $\sum_{k=0}^{21} \binom{21}{k} = 2^{21}$.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 $S + S = \sum_{k=0}^{21} \binom{21}{k} = 2^{21}$.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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