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

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MCQ

If C0 + C1 + C2 + ... + Cn = 256, then 2nC2 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:
\[2^n = 256\]
\[ \Rightarrow 2^n = 2^8 \]
\[ \Rightarrow n = 8\]
\[\therefore {}^{2n} C_2 = {}^{16} C_2 \]
\[ \Rightarrow^{16} C_2 = \frac{16!}{2! 14!} = \frac{16 \times 15}{2} = 120\]
Concept: Combination
  Is there an error in this question or solution?

APPEARS IN

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

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