Question

if nCr - 1 = 36,nCr = 84 and nCr+1 = 126, then find rC2. [NT : form eqn using nCr/nCr+1 and nCr/nCr-1 to find the value of r.]

Answer 1

Given, $^nC_{r-1}=36\\\\ ^nC_r=84\\\\ ^nC_{r+1}=126$

∴ $\frac{^nC_r}{^nC_{r-1}}=\frac{84}{36}$
⇒{n!/r!(n - r)!}/{n!/(r-1)!(n-r-1)!} = 7/3
⇒(n - r + 1)/r = 7/3
⇒3n - 3r + 3 = 7r
⇒3n + 3 = 10r ------(1)

Again, $\frac{^nC_{r+1}}{^nC_r}=\frac{126}{84}$
⇒{n!/(r+1)!(n-r-1)!}/{n!/r!(n-r)!} = 3/2
⇒(n - r)/(r + 1) = 3/2
⇒2n - 2r = 3r + 3
⇒2n - 3 = 5r ------(2)
From equations (1) and (2),
3n + 3 = 4n - 6
⇒9 = n and r = 3

Now, rC₂ = ³C₂ = 3!/2! = 3

Hence, answer is 3

Answer 2

Answer:

nCr-1=36,ncr=84,&ncr=84ncr+1=126

3x-10r=-3&4n-10r=6

on solving

n=9

n=3