Question

the number of non-negative intergal solutions of $x_1+x_2+x_3+4x_4=20$ is _______.

• A. $530$
• B. $532$
• C. $534$
• D. $536$

Answer 1

Answer: (D)

$536$

Consider Case$I$Let $x_4=0$

$\Rightarrow x_1+x_2+x_3=20$

we have to distribute $20$ identical objects into $3$ groups.

Number of solutions ${=}^{20+3-1}{C}_{3-1}{=}^{22}{C}_2$

Consider Case$II$Let $x_4=1$

$\Rightarrow x_1+x_2+x_3=20-4=16$

we have to distribute $16$ identical objects into $3$ groups.

Number of solutions ${=}^{16+3-1}{C}_{3-1}{=}^{18}{C}_2$

Consider Case$III$Let $x_4=2$

$\Rightarrow x_1+x_2+x_3=20-8=12$

we have to distribute $12$ identical objects into $3$ groups.

Number of solutions ${=}^{12+3-1}{C}_{3-1}{=}^{14}{C}_2$

Consider Case$IV$Let $x_4=3$

$\Rightarrow x_1+x_2+x_3=20-12=8$

we have to distribute $8$ identical objects into $3$ groups.

Number of solutions ${=}^{8+3-1}{C}_{3-1}{=}^{10}{C}_2$

Consider Case$V$Let $x_4=4$

$\Rightarrow x_1+x_2+x_3=20-16=4$

we have to distribute $4$ identical objects into $3$ groups.

Number of solutions ${=}^{4+3-1}{C}_{3-1}{=}^6{C}_2$

Consider Case$VI$Let $x_4=5$

$\Rightarrow x_1+x_2+x_3=20-20$

we have to distribute $0$ identical objects into $3$ groups.

Number of solutions ${=}^{0+3-1}{C}_{3-1}{=}^2{C}_2=1$

Hence total number of solutions$=1{+}^6{C}_2{+}^{10}{C}_2{+}^{14}{C}_2{+}^{18}{C}_2{+}^{22}{C}_2$

$=1+\frac{6!}{4!2!}+\frac{10!}{8!2!}+\frac{14!}{12!2!}+\frac{18!}{16!2!}+\frac{22!}{20!2!}$

$=1+\frac{6\times 5}{2}+\frac{10\times 9}{2}+\frac{14\times 13}{2}+\frac{18\times 17}{2}+\frac{22\times 21}{2}$

$=1+3\times 5+5\times 9+7\times 13+9\times 17+11\times 21$

$=1+15+45+91+153+231=536$