Question
The expression $$^n{C_r} + 4 \cdot {\,^n}{C_{r - 1}} + 6 \cdot {\,^n}{C_{r - 2}} + 4 \cdot {\,^n}{C_{r - 3}} + {\,^n}{C_{r - 4}}$$ is equal to
A.
$$ {\,^{n + 4}}{C_{r}}$$
B.
$$2 \cdot {\,^{n + 4}}{C_{r - 1}}$$
C.
$$4 \cdot {\,^n}{C_{r}}$$
D.
$$11 \cdot {\,^n}{C_{r}}$$
Answer :
$$ {\,^{n + 4}}{C_{r}}$$
Solution :
$$\eqalign{
& ^n{C_r} + 4 \cdot {\,^n}{C_{r - 1}} + 6 \cdot {\,^n}{C_{r - 2}} + 4 \cdot {\,^n}{C_{r - 3}} + {\,^n}{C_{r - 4}} \cr
& = \left( {^n{C_r} + {\,^n}{C_{r - 1}}} \right) + 3\left( {^n{C_{r - 1}} + {\,^n}{C_{r - 2}}} \right) + 3\left( {^n{C_{r - 2}} + {\,^n}{C_{r - 3}}} \right) + \left( {^n{C_{r - 3}} + {\,^n}{C_{r - 4}}} \right) \cr
& = {\,^{n + 1}}{C_r} + 3 \cdot {\,^{n + 1}}{C_{r - 1}} + 3 \cdot {\,^{n + 1}}{C_{r - 2}} + {\,^{n + 1}}{C_{r - 3}} \cr
& = \left( {^{n + 1}{C_r} + {\,^{n + 1}}{C_{r - 1}}} \right) + 2\left( {^{n + 1}{C_{r - 1}} + {\,^{n + 1}}{C_{r - 2}}} \right) + \left( {^{n + 1}{C_{r - 2}} + {\,^{n + 1}}{C_{r - 3}}} \right) \cr
& = {\,^{n + 2}}{C_r} + 2 \cdot \left( {^{n + 2}{C_{r - 1}} + {\,^{n + 1}}{C_{r - 2}}} \right) \cr
& = \left( {^{n + 2}{C_r} + {\,^{n + 2}}{C_{r - 1}}} \right) + \left( {^{n + 2}{C_{r - 1}} + {\,^{n + 2}}{C_{r - 2}}} \right) \cr
& = {\,^{n + 3}}{C_r} + {\,^{n + 3}}{C_{r - 1}} = {\,^{n + 4}}{C_r} \cr} $$