The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and …

Question

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

Accepted Answer

Choose $j$ identical items ($0\le j\le10$) and $10-j$ distinct items from the 21:

$$\sum_{j=0}^{10}\binom{21}{10-j} = \binom{21}{10}+\binom{21}{9}+\cdots+\binom{21}{0} = \frac{2^{21}}{2} = 2^{20}$$

(using the symmetry $\sum_{k=0}^{10}\binom{21}{k} = 2^{20}$ since $\binom{21}{k}$ for $k=0,\ldots,10$ accounts for exactly half of $2^{21}$)

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