Question

The number of non-negative integral solutions of $a+b+c+d=n,nϵN$, is

• A. ${}^{n+3}{P}_2$
• B. $\frac{(n+1)(n+2)(n+3)}{6}$
• C. ${}^{n-1}{C}_{n-4}$
• D. none of these

Answer 1

The correct option is B $\frac{(n+1)(n+2)(n+3)}{6}$

$a+b+c+d=n$
Now $nϵN$ and $a,b,c,dϵZ^{+}$
Consider $n=1$
We get
$a+b+c+d=1$
$a=0,b=0,c=0,d=1$
$a=0,b=0,c=1,d=0$
$a=0,b=1,c=0,d=0$
$a=1,b=0,c=0,d=0$
Hence we have 4 solutions
$={}^4{C}_3$ solutions
$={}^{1+3}{C}_3$ solutions
Let n=2
we get
$a=0,b=0,c=0,d=2$
$a=0,b=0,c=1,d=1$
$a=0,b=1,c=0,d=1$
$a=0,b=0,c=2,d=0$
$a=0,b=1,c=1,d=0$
$a=0,b=2,c=0,d=0$
$a=1,b=1,c=0,d=0$
$a=2,b=0,c=0,d=0$
$a=1,b=0,c=0,d=1$
$a=1,b=0,c=1,d=0$
Hence total 10 solutions
$={}^5{C}_3$
$={}^{2+3}{C}_3$
Hence for $a+b+c+d=n$ we will have ${}^{n+3}{C}_3$ solutions.
$=\frac{(n+3)(n+2)(n+1)}{3!}$
$=\frac{(n+3)(n+2)(n+1)}{6}$