Question

If $\sum\limits_{r = 1}^{25} {\left( {{}^{50}{C_r} \times {}^{50 - r}{C_{25 - r}}} \right)} = k\left( {{}^{50}{C_{25}}} \right)$, then k=?
A) ${2^{25}}$
B) ${25^2}$
C) ${2^{24}}$
D) ${2^{25}} - 1$

Answer 1

We can take the term inside the summation in the LHS. Then we can simplify it by expanding using the expansion of combinations. Then we can find the summation taking the sum by binomial expansion. Then we can compare it with the RHS and then solve for k.

Complete step by step solution:
We have the equation $\sum\limits_{r = 1}^{25} {\left( {{}^{50}{C_r} \times {}^{50 - r}{C_{25 - r}}} \right)} = k\left( {{}^{50}{C_{25}}} \right)$
We can take the term inside the summation.
Let ${P_r} = {}^{50}{C_r} \times {}^{50 - r}{C_{25 - r}}$ .
We know that the expansion of combinations is given by ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . So, the expression will become,
$\Rightarrow {P_r} = \dfrac{{50!}}{{r!\left( {50 - r} \right)!}} \times \dfrac{{\left( {50 - r} \right)!}}{{\left( {25 - r} \right)! \times \left( {50 - r - 25 + r} \right)!}}$
On simplification, we get,
$\Rightarrow {P_r} = \dfrac{{50!}}{{r!\left( {50 - r} \right)!}} \times \dfrac{{\left( {50 - r} \right)!}}{{\left( {25 - r} \right)! \times 25!}}$
On cancelling the common terms, we get,
$\Rightarrow {P_r} = \dfrac{{50!}}{{r! \times 25! \times \left( {25 - r} \right)!}}$
We can multiply both numerator and denominator with $25!$
$\Rightarrow {P_r} = \dfrac{{50! \times 25!}}{{r! \times 25! \times \left( {25 - r} \right)! \times 25!}}$
On rearranging, we get,
$\Rightarrow {P_r} = \dfrac{{50!}}{{{{\left( {25!} \right)}^2}}} \times \dfrac{{25!}}{{r!\left( {25 - r} \right)!}}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . So, the expression will become,
$\Rightarrow {P_r} = {}^{50}{C_{25}} \times {}^{25}{C_r}$
Now we can take the summation.
$\Rightarrow \sum\limits_{r = 1}^{25} {{P_r}} = \sum\limits_{r = 1}^{25} {{}^{50}{C_{25}} \times {}^{25}{C_r}}$
We can take the constant term outside the summation. So, we get,
$\Rightarrow \sum\limits_{r = 1}^{25} {{P_r}} = {}^{50}{C_{25}}\sum\limits_{r = 1}^{25} {{}^{25}{C_r}}$
Now we can add and subtract ${}^{25}{C_0}$ to the summation term. So, we get,
$\Rightarrow \sum\limits_{r = 1}^{25} {{P_r}} = {}^{50}{C_{25}} \times \left( {{}^{25}{C_0} + \sum\limits_{r = 1}^{25} {{}^{25}{C_r}} - {}^{25}{C_0}} \right)$
We know that ${a_0} + \sum\limits_{r = 1}^n {{a_r}} = \sum\limits_{r = 0}^n {{a_r}}$ . So, the summation will become,
$\Rightarrow \sum\limits_{r = 1}^{25} {{P_r}} = {}^{50}{C_{25}} \times \left( {\sum\limits_{r = 0}^{25} {{}^{25}{C_r}} - {}^{25}{C_0}} \right)$
We know that ${}^{25}{C_0} = 1$ and $\sum\limits_{r = 0}^n {{}^n{C_r}} = {2^n}$ . So, the expression will become,
$\Rightarrow \sum\limits_{r = 1}^{25} {{P_r}} = {}^{50}{C_{25}} \times \left( {{2^{25}} - 1} \right)$
Hence, we have,
$\Rightarrow \sum\limits_{r = 1}^{25} {\left( {{}^{50}{C_r} \times {}^{50 - r}{C_{25 - r}}} \right)} = {}^{50}{C_{25}} \times \left( {{2^{25}} - 1} \right)$
But it is given that $\sum\limits_{r = 1}^{25} {\left( {{}^{50}{C_r} \times {}^{50 - r}{C_{25 - r}}} \right)} = k\left( {{}^{50}{C_{25}}} \right)$ .
As the LHS are the same, we can equate the RHS.
$\Rightarrow k \times {}^{50}{C_{25}} = {}^{50}{C_{25}} \times \left( {{2^{25}} - 1} \right)$
On cancelling the common terms on both sides, we get,
$\Rightarrow k = \left( {{2^{25}} - 1} \right)$
Therefore, the required value of k is ${2^{25}} - 1$

So, the correct answer is option D.

Note:
We used the concept of combinations and its expansion to solve this problem. To find the sum of the combinations, we used the concept of binomial expansion. The sum of the combinations of zeroth to nth term is given by taking both the terms of the binomial as 1. Before taking the sum using binomial expansion, we must make sure that the summation is from zero to n. For making it zero ton, we add and subtract the zeroth term and modify the function accordingly.