# If C0 + C1 + C2 + ... + Cn = 256, Then 2nc2 is Equal to (A) 56 (B) 120 (C) 28 (D) 91 - Mathematics

MCQ

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

• 56

• 120

• 28

• 91

#### 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:
$2^n = 256$
$\Rightarrow 2^n = 2^8$
$\Rightarrow n = 8$
$\therefore {}^{2n} C_2 = {}^{16} C_2$
$\Rightarrow^{16} C_2 = \frac{16!}{2! 14!} = \frac{16 \times 15}{2} = 120$
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Q 17 | Page 26