Question

The number of values of $r$ satisfying the equation ${}^{39}{C}_{3r-1}-{}^{39}{C}_{r^2}={}^{39}{C}_{r^2-1}-{}^{39}{C}_{3r}$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

${{}^{39}C}_{3r-1}-{{}^{39}C}_{r^2}={{}^{39}C}_{r^2-1}-{{}^{39}C}_{3r}$

As ${{}^{n}C}_r+{{}^{n}C}_{r+1}={{}^{n+1}C}_{r+1}$

So, ${{}^{39}C}_{3r-1}-{{}^{39}C}_r={{}^{39}C}_{r^2-1}-{{}^{39}C}_{3r^2}$

$\Rightarrow {{}^{40}C}_{3r}={{}^{40}C}_{r^2}$

here either $3r-r^2$ or $40-3r=r^2$

So, for $3r=r^2$ $r^2+3r-40=0$

$r=3,0$ $r=5,-8$

As $r$ can't be $-8$ and $0$ because ${{}^{39}C}_{3r-1}$ is not define.

Hence, the answer is two values of $r=3,5.$