The number of positive integral solutions of the inequality x+y+z≤ 20 is

Question

The number of positive integral solutions of the inequality x+y+z≤ 20 is (A) 1008 (B) 1028 (C) 1108 (D) 1140

Tardigrade Admin · Accepted Answer

The number of solutions (positive integer) of

x + y + z = r

is

^{r - 1} C_{2}

∴

The desired number of solutions is

\sum_{r = 3}^{20} r^{r - 1} C_{2}

=^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots\dots . +^{19} C_{2}

= (^{3} C_{3} +^{3} C_{2}) +^{4} C_{2} + \dots\dots +^{19} C_{2}

=^{4} C_{3} +^{4} C_{2} + \dots\dots . +^{18} C_{2} +^{19} C_{2}

=^{20} C_{3}

(Adding in the same way)

= \frac{20 \times 19 \times 183}{3 \times 2} = 1140

Q. The number of positive integral solutions of the inequality $x + y + z \leq 20$ is