Question

The sum \[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} \] for \[i \in \left[ {0,m} \right]\] where \[\left( {\begin{array}{*{20}{c}}
  p \\
  q
\end{array}} \right) = 0\] if \[p > q\] is maximum when \[m\] is
A. 5
B. 10
C. 15
D. 20

Answer 1

Hint:In this question, we need to find the value of \[m\]. For this, we have to use the formula of combination to determine the value of \[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} \]. After that, we will find the value of \[m\] using the condition such as \[\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)\] is maximum when \[r = \dfrac{n}{2}\].

Formula used:
The combination formula used for selecting \[r\] things from \[n\] things is given by
\[\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step-by-step answer:
We know that \[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} \]
Here, \[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} \] is the coefficient of \[{x^m}\] in the expansion of \[{\left( {1 + x} \right)^{10}}{\left( {x + 1} \right)^{20}}\]
Let us simplify this first.
\[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} = \sum\limits_{i = 0}^m {{}^{10}{C_i} \times {}^{20}{C_{m - i}}} \]
\[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} = {}^{10}{C_0} \times {}^{20}{C_m} + {}^{10}{C_1} \times {}^{20}{C_{m - 1}} + {}^{10}{C_2} \times {}^{20}{C_{m - 2}} + ..... + {}^{10}{C_m}\]
Thus, it gives
\[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} = {}^{30}{C_m}.....\] using the above concept.
Also, we can say that \[\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)\] is maximum when \[{\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)_{\max }}\left\{ {\begin{array}{*{20}{c}}
  {r = \dfrac{n}{2}.....n \in {\text{even}}} \\
  {r = \dfrac{{n + 1}}{2}.......n \in odd}
\end{array}} \right.\]
Here, \[n\] is even.
So, \[{\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)_{\max }} \Rightarrow r = \dfrac{n}{2}\]
\[{\left( {\begin{array}{*{20}{c}}
  {30} \\
  m
\end{array}} \right)_{\max }} \Rightarrow m = \dfrac{{30}}{2}\]
Thus, we get \[{\left( {\begin{array}{*{20}{c}}
  {30} \\
  m
\end{array}} \right)_{\max }} \Rightarrow m = 15\]
Hence, the value of \[m\] is 15.

Therefore, the correct option is (C)

Note: Many students make mistakes in the simplification part and also, in using the combination formula. We need to use appropriate exponential properties while solving this question. We have to use the concept of summation in an appropriate way while simplifying \[\sum\limits_{i = 0}^m {\left( {\begin{array}{*{20}{c}}
  {10} \\
  i
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  {20} \\
  {m - i}
\end{array}} \right)} \] part.