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,

^{n}_{Cr} + ^{n}C_{r+1} + ^{n}C_{r+1} + ^{n}C_{r+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+2}C_{r+2}

= C(n + 2 , r + 2)