If C0-C1-C2-Cn-256- then 2 n C2 is equal toa 56b 120c 28d 91

(b) 150 If set S nbsp;has n elements, then nbsp;C #160;n, #160;k nbsp;is the number of ways of choosing k elements from S. Thus, the number of subset ...

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Byju's Answer

Standard XI

Mathematics

Combination with Restrictions

If C0+C1+C2+…...

Question

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) $91$

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Solution

Correct option is (b) $120$
We know that,
$(x + a)^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} + . . . . +^{n} C_{n} a^{n}$
Putting, $x = a = 1$ , we get,
$\Rightarrow 2^{n} = C_{0} + C_{1} + C_{2} + . . . + C_{n}$
$\Rightarrow 2^{n} = 256$
$\Rightarrow 2^{n} = 2^{8}$
$∴ n = 8$
Hence, $^{2 n} C_{n} =^{16} C_{8}$
$= \frac{16!}{8! 8!}$
$= \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}$
$= 120$

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If C0+C1+C2+...+Cn=256, then 2nCn is equal to(a) 56(b) 120(c) 28(d) 91

Correct option is (b) 120We know that,(x+a)n=nC0xn+nC1xn−1a1+nC2xn−2a2+....+nCnanPutting, x=a=1, we get,⇒2n=C0+C1+C2+...+Cn⇒2n=256⇒2n=28∴n=8Hence, 2nCn=16C8=16!8!8!=16×15×14×13×12×11×10×98×7×6×5×4×3×2=120

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) $91$