Question

The sum $\sum\limits_{i=0}^{m}{\left( \begin{matrix} 10 \\\ i \\\ \end{matrix} \right)\left( \begin{matrix} 20 \\\ m-i \\\ \end{matrix} \right)}$ be maximum when m is
(a) 15
(b) 5
(c) 10
(d) 20

Answer 1

First we have to expand the series given as $\sum\limits_{i=0}^{m}{\left( \begin{matrix} 10 \\\ i \\\ \end{matrix} \right)\left( \begin{matrix} 20 \\\ m-i \\\ \end{matrix} \right)}$ . So, on expanding we will get as ${}^{10}{{C}_{0}}\cdot {}^{20}{{C}_{m}}+{}^{10}{{C}_{1}}\cdot {}^{20}{{C}_{m-1}}+....+{}^{10}{{C}_{m}}\cdot {}^{20}{{C}_{0}}$ . Then, we have to consider ${}^{10}{{C}_{i}}$ as any colour balls and ${}^{20}{{C}_{m}}$ as any colour balls and then taking average of both balls. We will get a value of m.

Complete step-by-step answer:
In the question we are given equation $\sum\limits_{i=0}^{m}{\left( \begin{matrix} 10 \\\ i \\\ \end{matrix} \right)\left( \begin{matrix} 20 \\\ m-i \\\ \end{matrix} \right)}$ and we have to find at which value of m the summation will be maximum.
So, expanding the terms we get
${}^{10}{{C}_{0}}\cdot {}^{20}{{C}_{m}}+{}^{10}{{C}_{1}}\cdot {}^{20}{{C}_{m-1}}+....+{}^{10}{{C}_{m}}\cdot {}^{20}{{C}_{0}}$ …………………………..(1)
Here, we will assume that ${}^{10}{{C}_{i}}$ as green balls and ${}^{20}{{C}_{m}}$ as red balls. So, here maximum 10 green balls and 20 red balls are there
Therefore, we have to choose m number of balls So, that sum will be maximum. So, we will used average formula which will be
$m=\dfrac{green\ balls+red\ balls}{2}=\dfrac{10+20}{2}=15$
Thus, the value of m is 15.
Hence, option (a) is correct.

So, the correct answer is “Option A”.

Note: Another method to find value of m is by taking option method. Suppose taking m as 5 and substituting in the terms ${}^{10}{{C}_{0}}\cdot {}^{20}{{C}_{m}}+{}^{10}{{C}_{1}}\cdot {}^{20}{{C}_{m-1}}+....+{}^{10}{{C}_{m}}\cdot {}^{20}{{C}_{0}}$ and adding all the terms will be time consuming. This method we have to perform by taking the value of m as 5, 10, 15, 20 and then comparing and among them the highest value of summation will be m. Though answers will be obtained but it will become tedious and a waste of time. So, ignore this method.