Question

If ${(}^{15}{C}_r{+}^{15}{C}_{r-1}\left){(}^{15}{C}_{15-r}{+}^{15}{C}_{16-r}\right)={(}^{16}{C}_{13}{)}^2$, then the value of $r$ is

• A. $r=2$
• B. $r=3$
• C. $r=4$
• D. none of these

Answer 1

Answer: (B)

$r=3$

${(}^{15}{C}_r{+}^{15}{C}_{r-1}\left){(}^{15}{C}_{15-r}{+}^{15}{C}_{16-r}\right)={(}^{16}{C}_{13}{)}^2$
${}^{16}{C}_r{(}^{15}{C}_r{+}^{15}{C}_{r-1})={(}^{16}{C}_{13}{)}^2$ (${\because }^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$)
${}^{16}{C}_r{(}^{16}{C}_r)={(}^{16}{C}_3{)}^2$ (${\because }^n{C}_r{=}^n{C}_{n-r}$)
${(}^{16}{C}_r{)}^2={(}^{16}{C}_3{)}^2$
$\Rightarrow r=3$