Question
The number of non-negative integral solutions of $$a + b + c + d = n,n \in N,$$ is
A.
$$^{n + 3}{P_2}$$
B.
$$\frac{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{6}$$
C.
$$^{n - 1}{C_{n - 4}}$$
D.
None of these
Answer :
$$\frac{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{6}$$
Solution :
The number of non-negative integral solutions
= co-efficient of $${x^n}\,{\text{in }}{\left( {{x^0} + {x^1} + {x^2} + .....} \right)^4}$$
= co-efficient of $${x^n}\,{\text{in }}{\left( {1 - x} \right)^{ - 4}}$$
= co-efficient of $${x^n}\,{\text{in }}\left\{ {^3{C_0} + {\,^4}{C_1}x + {\,^5}{C_2}{x^2} + .....} \right\}$$
$$ = {\,^{n + 3}}{C_n} = \frac{{\left( {n + 3} \right)!}}{{n!\,\,3!}} = \frac{{\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right)}}{6}.$$