Question

The value of nCn - 1 + nCn - 2 is(a) n/2(b) n(n + 1)/2(c) n(n-1)/2(d) (n + 1)²/2​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle\sf{\,^{n}C_{n-1}+\,^{n}C_{n-2}}$

$\displaystyle\sf{=\dfrac{n!}{(n-1)!(n-(n-1))!}+\dfrac{n!}{(n-2)!(n-(n-2))!}}$

$\displaystyle\sf{=\dfrac{n!}{(n-1)!(n-n+1)!}+\dfrac{n!}{(n-2)!(n-n+2)!}}$

$\displaystyle\sf{=\dfrac{n!}{(n-1)!(1)!}+\dfrac{n!}{(n-2)!(2)!}}$

$\displaystyle\sf{=\dfrac{n(n-1)!}{(n-1)!(1)!}+\dfrac{n(n-1)(n-2)!}{(n-2)!(2)!}}$

$\displaystyle\sf{=\dfrac{n}{1}+\dfrac{n(n-1)}{2}}$

$\displaystyle\sf{=n+\dfrac{n^2-n}{2}}$

$\displaystyle\sf{=\dfrac{2n+n^2-n}{2}}$

$\displaystyle\sf{=\dfrac{n^2+n}{2}}$

$\displaystyle\sf{=\dfrac{n(n+1)}{2}}$