Let $n_{1}

Question

Let $n_{1}<n_{2}<n_{3}<n_{4}<n_{5}$ be positive integers such that $n_{1}$ $+n_{2}+n_{3}+n_{4}+n_{5}=20$. Then the number of such distinct arrangements $\left(n_{1}, n_{2}, n_{3}, n_{4}, n_{5}\right)$ is

Professor PC Thomas Classes · Answer

(7): As $n_{1} \geq 1, n_{2} \geq 2$ etc. We have $a+b+c+d+e$\ $=20-(1+2+3+4+5)=5$\ i.e. $a+b+c+d+e=5$ Now $a \leq b \leq c \leq d \leq e$. So the cases can be listed giving\ (0,0,0,0,5);(0,0,0,1,4);(0,0,0,2,3);(0,0,1,1,3); ( $0,0,1,2,2$ ) ; ( $0,1,1,1,2$ ); ( $1,1,1,1,1$ )\ So there are 7 distinct arrangements.\ Alternative solution: The largest $e$ is 10 . (given)\ $a b c d e$\ 123410\ and least $e$ is 6 .\ 2, 3, 4, 5, 6\ So the number of ways\ $={ }^{4} C_{0}+\frac{{ }^{4} C_{1}}{4}+\frac{{ }^{4} C_{2}}{3}+\frac{{ }^{4} C_{3}}{2}+\frac{{ }^{4} C_{4}}{1}$\ $=1+1+2+2+1=7$\