If sumlimits_{r = 0}^{25} {left( {^{50}C_r cd | vedclass

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(D) We know that $^{n}C_{r} \cdot ^{n-r}C_{k-r} = \frac{n!}{r!(n-r)!} \cdot \frac{(n-r)!}{(k-r)!(n-k)!} = \frac{n!}{r!(k-r)!(n-k)!}$.Multiplying and d

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$2^{25}$

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(D) We know that $^{n}C_{r} \cdot ^{n-r}C_{k-r} = \frac{n!}{r!(n-r)!} \cdot \frac{(n-r)!}{(k-r)!(n-k)!} = \frac{n!}{r!(k-r)!(n-k)!}$.Multiplying and dividing by $k!$,we get $\frac{n!}{k!(n-k)!} \cdot \frac{k!}{r!(k-r)!} = ^{n}C_{k} \cdot ^{k}C_{r}$.Applying this to the given sum:$\sum\limits_{r = 0}^{25} {^{50}C_r \cdot ^{50 - r}C_{25 - r}} = \sum\limits_{r = 0}^{25} {^{50}C_{25} \cdot ^{25}C_r}$.$= ^{50}C_{25} \sum\limits_{r = 0}^{25} {^{25}C_r}$.Since $\sum\limits_{r = 0}^{n} {^{n}C_r} = 2^n$,we have $\sum\limits_{r = 0}^{25} {^{25}C_r} = 2^{25}$.Therefore,the expression becomes $^{50}C_{25} \cdot 2^{25}$.Comparing this with $K\left( {^{50}C_{25}} \right)$,we get $K = 2^{25}$.

If $\sum\limits_{r = 0}^{25} {\left( {^{50}C_r \cdot ^{50 - r}C_{25 - r}} \right) = K\left( {^{50}C_{25}} \right)}$,then $K$ is equal to

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