Question

The value of $\left(\begin{array}{c}50\\ 0\end{array}\right)\left(\begin{array}{c}50\\ 1\end{array}\right)+\left(\begin{array}{c}50\\ 1\end{array}\right)\left(\begin{array}{c}50\\ 2\end{array}\right)+........+\left(\begin{array}{c}50\\ 49\end{array}\right)\left(\begin{array}{c}50\\ 50\end{array}\right)$ is , where ${}^n{C}_r=\left(\begin{array}{c}n\\ r\end{array}\right)$

• A. $\left(\begin{array}{c}100\\ 50\end{array}\right)$
• B. $\left(\begin{array}{c}100\\ 51\end{array}\right)$
• C. $\left(\begin{array}{c}50\\ 25\end{array}\right)$
• D. ${\left(\begin{array}{c}50\\ 25\end{array}\right)}^2$

Answer 1

The correct option is B $\left(\begin{array}{c}100\\ 51\end{array}\right)$

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+......{+}^n{C}_n{x}^n$.

Now, $(1+x{)}^{2n}=(1+x{)}^n.(1+x{)}^n$

$=({}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+.....{.}^n{C}_{n-1}{x}^{n-1}{+}^n{C}_n{x}^n).({}^n{C}_n{x}^n{+}^n{C}_{n-1}{x}^{n-1}+......{+}^n{C}_2{x}^2{+}^n{C}_{1}x{+}^n{C}_0)$.

Now, equating the co-efficient of $x^{n-1}$ both sides we get,

${}^{2n}{C}_{n-1}={(}^n{C}_0\left){(}^n{C}_{n-1}\right)+{(}^n{C}_1\left){(}^n{C}_{n-2}\right)+......{(}^n{C}_{n-1}\left){(}^n{C}_0\right)$

or, ${}^{2n}{C}_{n+1}={(}^n{C}_0\left){(}^n{C}_1\right)+{(}^n{C}_1\left){(}^n{C}_2\right)+......{(}^n{C}_{n-1}\left){(}^n{C}_n\right)$. {Since ${}^n{C}_r{=}^n{C}_{n-r}$]

Now, putting $n=50$ we get ,

${}^{100}{C}_{51}=$${(}^{50}{C}_0\left){(}^{50}{C}_1\right)+{(}^{50}{C}_1\left){(}^{50}{C}_2\right)+......{(}^{50}{C}_{49}\left){(}^{50}{C}_{50}\right)$.