Question

If ${}^{15}{P}_{r-1}{:}^{15}{P}_{r-2}=3:4$, then $r=$

• A. $\frac{65}{4}$
• B. $\frac{55}{4}$
• C. $\frac{71}{4}$
• D. $\frac{67}{4}$

Answer 1

Answer: (A)

$\frac{65}{4}$

Given that,

${}^{15}{P}_{r-1}:{}^{15}{P}_{r-2}=3:4$

$\frac{{}^{15}{P}_{r-1}}{{}^{15}{P}_{r-2}}=\frac{3}{4}$

We know that

${}^n{P}_r=\frac{n!}{(n-r)!}$

Therefore,

$\frac{{}^{15}{P}_{r-1}}{{}^{15}{P}_{r-2}}=\frac{3}{4}$

$\frac{\frac{15!}{(15-r+1)!}}{\frac{15!}{(15-r+2)!}}=\frac{3}{4}$

$\frac{(17-r)!}{(16-r)!}=\frac{3}{4}$

$\frac{(17-r)(17-r-1)!}{(16-r)!}=\frac{3}{4}$

$\frac{(17-r)(16-r)!}{(16-r)!}=\frac{3}{4}$

$17-r=\frac{3}{4}$

$68-4r=3$

$4r=65$

$r=\frac{65}{4}$

Hence, this is the answer.