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

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(A) We know that $^{n}{C_r} \cdot ^{n-r}{C_k} = ^{n}{C_k} \cdot ^{n-k}{C_{r}}$.Applying this identity to the given expression:$^{50}{C_r} \cdot ^{50-r

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$2^{25}$

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(A) We know that $^{n}{C_r} \cdot ^{n-r}{C_k} = ^{n}{C_k} \cdot ^{n-k}{C_{r}}$.Applying this identity to the given expression:$^{50}{C_r} \cdot ^{50-r}{C_{25-r}} = ^{50}{C_{25}} \cdot ^{50-25}{C_{r}} = ^{50}{C_{25}} \cdot ^{25}{C_r}$.Now,substitute this into the summation:$\sum\limits_{r = 0}^{25} {\left( {^{50}{C_r} \cdot ^{50 - r}{C_{25 - r}}} \right)} = \sum\limits_{r = 0}^{25} {\left( {^{50}{C_{25}} \cdot ^{25}{C_r}} \right)}$.Since $^{50}{C_{25}}$ is independent of $r$,we can take it out of the summation:$= ^{50}{C_{25}} \sum\limits_{r = 0}^{25} {^{25}{C_r}}$.We know that $\sum\limits_{r = 0}^{n} {^{n}{C_r} = 2^n}$. Therefore,$\sum\limits_{r = 0}^{25} {^{25}{C_r} = 2^{25}}$.Thus,the expression becomes $^{50}{C_{25}} \cdot 2^{25}$.Comparing this with $K \cdot ^{50}{C_{25}}$,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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