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What is C(n, r) + 2C(n, r + 1) + C(n, r + 2) equal to?
1. C(n + 1, r)
2. C(n + 1, r + 2)
3. C(n + 2, r + 2)
4. C(n + 2, r + 3)

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Correct Answer - Option 3 : C(n + 2, r + 2)

Concept:

\(\rm C(n,r)=\frac{n!}{(n-r)!r!}\)

Calculation: 

The given problem can be written as,

nCr + nCr+1 + nCr+1 + nCr+2

\(\rm \frac{n!}{r!(n - r)!)} + \frac{n!}{(r+1)!(n - r-1)!)} + \frac{n!}{(r+1)!(n - r-1)!)} + \frac{n!}{(r+2)!(n - r-2)!)}\)

=\(\rm \frac{n!}{r!(n - r-1)!)} [\frac{1}{(n-r)}+\frac{1}{(r+1)}]+\frac{n!}{(r+1)!(n - r-2)!)}[\frac{1}{(n-r-1)}+\frac{1}{(r+2)}]\)

=\(\rm \rm \frac{n!}{r!(n - r-1)!)} [\frac{r+1+n-r}{(n-r)(r+1)}]+\frac{n!}{(r+1)!(n - r-2)!)}[\frac{r+2+n-r-1}{(n-r-1)(r+2)}]\)

=\(\rm \frac{n!\times (n+1)}{r!\times(r+1)\times(n-r-1)\times(n-r)} + \frac{n!}{(r+1)!(n-r-2)!}\times \frac{(n+1)}{(n-r-1)(r+2)}\)

\(\rm= \frac{(n+1)!}{(r+1)!(n-r)!}+\frac{n!\times(n+1)}{(r+2)(r+1)\times(n-r-1)(n-r-2)}\)

=\(\rm \frac{(n+1)!}{(r+1)!(n-r)!}+\frac{(n+1)!}{(r+2)!(n-r-1)!}\)

=\(\rm \frac{(n+1)!}{(r+1)!(n-r-1)!}[\frac{1}{n-r}+\frac{1}{r+2}]\)

=\(\rm \frac{(n+1)!}{(r+1)!(n-r-1)!}[\frac{n-r+r+2}{(r+2)(n-r)}]\)

=\(\rm \frac{(n+2)(n+1)!}{(r+2)(r+1)!(n-r)(n-r-1)!}\)

=\(\rm \frac{(n+2)!}{(r+2)!(n-r)!}\)

=n+2Cr+2

= C(n + 2 , r + 2)

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