Question

The median of $^{2{{n}}}{{{C}}_0}{,^{2{{n}}}}{{{C}}_1}{,^{2{{n}}}}{{{C}}_2}, \ldots .{,^{2{{n}}}}{{{C}}_{{n}}}$ (when ${{n}}$is even) is:
A. $^{2n}{C_{\frac{n}{2}}}$
B. $^{2n}{C_{\frac{{n + 1}}{2}}}$
C. $^{2n}{C_{\frac{{n - 1}}{2}}}$
D. None of these

Answer 1

Hint: We will use the concept of binomial coefficient to solve the question then we will first check the number of terms whether they are even or odd. After that we will apply the formula of median and substitute the values in the formula and get the answer.

Formula Used:
The median of any series can be calculated with the help of the formula: $\left( {\dfrac{{{n}}}{2} + 1} \right)$ th term

Complete step by step solution:
We have been given the observations $^{2{{n}}}{{{C}}_0}{,^{2{{n}}}}{{{C}}_1}{,^{2{{n}}}}{{{C}}_2}, \ldots .{,^{2{{n}}}}{{{C}}_{{n}}}$
Here number of terms $ = n + 1\quad $....(odd)
Median $ = \left( {\dfrac{{{n}}}{2} + 1} \right)$ th term
  ${ = ^{2n}}{C_{\frac{n}{2}}}$

Option ‘A’ is correct

Note: binomial coefficient states that he coefficient of ${x^k}$ in the expansion of ${(1 + x)^n}$ can be thought of as a binomial coefficient, or C(n, k). Informally, the number of k-element subsets (or k-combinations) of an n-element set may be expressed as the number of ways, disregarding order, in which k items can be selected from among n objects using a binomial coefficient C(n, k).