$^{m} C_{r} +^{m} C_{r - 1} .^{n} C_{1} +^{m} C_{r - 2},^{n} C_{2} + . . . +^{m} C_{1} .^{n} C_{r - 1} +^{n} C_{r} =$

^{m + n} C_{r - 1}

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^{m + n} C_{r}

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^{m + n} C_{r + 1}

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none of the above

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Solution

The correct option is A $^{m + n} C_{r}$
The given series
$=^{m} {C_{r} .}^{n} C_{0} +^{m} C_{r - 1} .^{n} C_{1} + . . . +^{m} C_{0} .^{n} C_{r},$
Now, ${(1 + x)}^{m} =^{m} C_{0} +^{m} C_{1} . x +^{m} C_{2} . x^{2} + . . . +^{m} C_{r} . x^{r} + . . . +^{m} C_{m} x^{n},$
and, ${(1 + n)}^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . +^{n} {C_{r} x^{r} + . . . +}^{n} C_{n} x^{n}$
The given series is the coefficient of $x^{r}$ in the product of $R . H . S .$ of above two.
$∴$ Sum of the series $=$ coefficient of $x^{r}$ in ${(1 + x)}^{m} . {(1 + x)}^{n}$
$=$ coefficient of $x^{r}$ in ${(1 + x)}^{m + n}$
$=^{m + n} C_{r} .$

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