If $C_{0} + C_{1} + C_{2} + . . . . + C_{n} = 256$ , then $^{2 n} C_{2}$ is equal to

56

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120

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28

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91

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Solution

The correct option is D $120$
We know that
$\Rightarrow {(x + a)}^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} a +^{n} C_{2} x^{n - 2} a . . . . . . .^{n} C_{n} a^{n}$
$P u t x = a = 1 \Rightarrow 2^{n} = C_{0} + C_{1} + C_{2} + . . . . . . C_{n} \Rightarrow 2^{n} = 256 \Rightarrow 2^{n} = 2^{8}$

$w e g e t, n = 8 t h e n^{2 n} C_{n} =^{16} C_{8} = \frac{16, 15, 14, 13, 12, 11, 10, 9, 8!}{8, 7, 6, 5, 4, 3, 2, 1, 8!}$
$H e n c e = 120$

So, option $B$ is correct answer.

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Similar questions

If $C_{0} + C_{1} + C_{2} + . . . + C_{n} = 256$ = then $^{2 n} C_{2}$ is equal to

If $C_{0} + C_{1} + C_{2} + . . . + C_{n} = 256$ , then $^{2 n} C_{n}$ is equal to
(a) $56$
(b) $120$
(c) $28$
(d) $91$

C_{0}, C_{1}, C_{2} . . . . . . . . . . ., C_{n}

(1 + x)^{n}, n

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . + C_{n - 1})

C_{0}, C_{1}, C_{2}, . . . . . . . . . . C_{n}

(1 + x)^{n}

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . . + C_{n - 1})

{(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . . . + C_{n} x^{R}

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1}

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