Question

If C₀ + C₁ + C₂ + ... + C_n = 256, then ²ⁿC₂ is equal to

Options

• 56

• 120

• 28

• 91

Answer 1

120

If set \[S\] has n elements, then 

\[C \left( n, k \right)\]  is the number of ways of choosing k elements from \[S\]

Thus, the number of subsets of  \[S\] of all possible values is given by

\[C\left( n, 0 \right) + C\left( n, 1 \right) + C\left( n, 3 \right) + . . . + C\left( n, n \right) = 2^n\]

Comparing the given equation with the above equation:

\[2^n = 256\]
\[ \Rightarrow 2^n = 2^8 \]
\[ \Rightarrow n = 8\]

\[\therefore {}^{2n} C_2 = {}^{16} C_2 \]
\[ \Rightarrow^{16} C_2 = \frac{16!}{2! 14!} = \frac{16 \times 15}{2} = 120\]