Solveeit Logo
Login

Question

Mathematics Question on permutations and combinations

If nCr−1=28, nCr=56^nC_{r-1}=28,\,^nC_r =56 and nCr+1=70^nC_{r+1}=70, then r=r =

A

11

B

22

C

33

D

44

Answer

33

Explanation

Solution

nCrnCr−1=5628\frac{^{n}C_{r}}{^{n}C_{r-1}}=\frac{56}{28} ⇒n!r!(n−r)!⋅(r−1)!(n−r+1)!n!=2\Rightarrow \frac{n!}{r ! \left(n-r\right)!}\cdot\frac{\left(r-1\right)!\left(n-r+1\right)!}{n!}=2 ⇒n−r+1r=2\Rightarrow \frac{n-r+1}{r}=2 ⇒n−r+1=2r\Rightarrow n-r + 1 = 2 r ⇒n=3r−1...(1)\Rightarrow n=3r-1 ...\left(1\right) Again nCr+1nCr=7056 \frac{^{n}C_{r+1}}{^{n}C_{r}}=\frac{70}{56} ⇒n!(r+1)!(n−r−1)!\Rightarrow\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!} r!(n−r)!n!=54\frac{r!\left(n-r\right)!}{n!}=\frac{5}{4} ⇒n−rr+1=54\Rightarrow \frac{n-r}{r+1}=\frac{5}{4} ⇒4n−4r=5r+5\Rightarrow 4n-4r=5r+5 ⇒4n=9r+5\Rightarrow 4n=9r+5 From (1)\left(1\right) and (2)\left(2\right), we get 12r−4=9r+512r - 4 = 9r + 5 ⇒3r=9\Rightarrow 3r=9 ⇒r=3\Rightarrow r=3.