The value of [C(7, 0) + C(7, 1)] + [C(7, 1) + C(7, 2)] + … + [C(7, 6) + C(7, 7)] is

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NDA (Held On: 23 April 2017) Maths Previous Year paper

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Answer 1

Concept:

C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) ………….. C(n, n) = 2ⁿ
C(n, 0) = C(n, n) = 1
C(n, r) = \(^{\rm{n}}{{\rm{C}}_{\rm{r}}} = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)

Calculation:

Given that,

⇒C(7, 0) + C(7, 1) + C(7, 1) + C(7, 2) + C(7, 2) … + C(7, 6) + C(7, 7)

⇒ C(7, 0) + 2.[ C(7, 1) + C(7, 2) + … + C(7, 6)] + C(7, 7)

Add and subtract C(7, 0),C(7, 7) so,

⇒C(7, 0) + 2.[ C(7, 1) + C(7, 2) + … + C(7, 6)] + C(7, 0) + C(7, 7) + C(7, 0) + C(7, 7) - C(7, 0) - C(7, 7)

⇒ 2[ C(7, 0) + C(7, 1) + C(7, 2) + … + C(7, 6) + C(7, 7)] – C(7,0) – C(7, 7)

⇒2[2⁷] - C(7,0) – C(7, 7) ...[∵ C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) ………….. C(n, n) = 2ⁿ]

⇒2 × 128 – 1 – 1

⇒ 254