MCQ

If *C*_{0} + *C*_{1} + *C*_{2} + ... + *C _{n}* = 256, then

^{2}

^{n}C_{2}is equal to

#### Options

56

120

28

91

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#### Solution

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:

\[2^n = 256\]

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

\[ \Rightarrow n = 8\]

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

\[ \Rightarrow n = 8\]

\[\therefore {}^{2n} C_2 = {}^{16} C_2 \]

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

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

Concept: Concept of Combinations

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