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

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SalmonWorldTopper SalmonWorldTopper · Answer 1 · 2021-10-08T14:48:14+02:00

Step-by-step explanation:

We have,

[tex]\displaystyle\sf{\,^{n}C_{n-1}+\,^{n}C_{n-2}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n!}{(n-1)!(n-(n-1))!}+\dfrac{n!}{(n-2)!(n-(n-2))!}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n!}{(n-1)!(n-n+1)!}+\dfrac{n!}{(n-2)!(n-n+2)!}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n!}{(n-1)!(1)!}+\dfrac{n!}{(n-2)!(2)!}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n(n-1)!}{(n-1)!(1)!}+\dfrac{n(n-1)(n-2)!}{(n-2)!(2)!}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n}{1}+\dfrac{n(n-1)}{2}}[/tex]

[tex]\displaystyle\sf{=n+\dfrac{n^2-n}{2}}[/tex]

[tex]\displaystyle\sf{=\dfrac{2n+n^2-n}{2}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n^2+n}{2}}[/tex]

[tex]\displaystyle\sf{=\dfrac{n(n+1)}{2}}[/tex]

joaobezerra joaobezerra · Answer 2 · 2022-03-15T12:43:11+01:00

Using the combination formula, it is found that the value of the expression is given by:

(b) n(n + 1)/2

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem:

[tex]C_{n, n-1} = \frac{n!}{(n-1)!(n - n + 1)!} = \frac{n(n-1)!}{(n-1)!} = n[/tex]

[tex]C_{n, n-2} = \frac{n!}{(n-2)!(n - n + 2)!} = \frac{n(n-1)(n-2)!}{(n-2)!2!} = \frac{n(n-1)}{2}[/tex]

Then, adding these values:

[tex]n + \frac{n(n-1)}{2} = \frac{2n + n^2 - n}{2} = \frac{n(n + 1)}{2}[/tex]

Hence option b is correct.

More can be learned about the combination formula at https://brainly.com/question/25821700