Question

If $a,b,c,d$ are odd natural numbers such that $a+b+c+d=20$ then the number of values of the ordered quadruplet $(a,b,c,d)$ is

• A. $165$
• B. $455$
• C. $310$
• D. none of these

Answer 1

Answer: (A)

$165$

let, $a=2n_1+1,b=2n_2+1,c=2n_3+1,d=2n_4+1$ where $n_i\in N$
$⟹2n_1+1+2n_2+1+2n_3+1+2n_4+1=20$
$⟹2n_1+2n_2+2n_3+2n_4=16$
$⟹n_1+n_2+n_3+n_4=8$
Number of solutions or ordered pairs for this equation are given by ${}^{n+r-1}{\text{C}}_{r-1}$
$={}^{8+4-1}{\text{C}}_{4-1}$
$={}^{11}{\text{C}}_3$
$=165$