Question

If $\rm ^{n}C_{r-1}=495$ , $\rm ^{n}C_{r}=220$ , $\rm ^{n}C_{r+1}=66$ find $\rm ^{r}C_{4}$.

Answer 1

Answer:

126

Step-by-step explanation:

Hi,

Given ⁿCr₋1 = 495

ⁿCr = 220,

ⁿCr+1 = 66

We know that ⁿCr  = n!/(n-r)!r!

ⁿCr₋1 = 495

n!/(n-r+1)!(r-1)! = 495-----(1)

ⁿCr = 220,

n!/(n-r)!r!  = 220 ---------(2)

ⁿCr+1 = 66

n!/(n-r-1)!(r+1)! = 66-------(3)

Dividing (1)/(2) we get

[n!/(n-r+1)!(r-1)!]/[n!/(n-r)!r!] = 495/220

⇒r/(n-r+1) = 9/4

⇒4r = 9n - 9r + 9

⇒13r - 9n = 9--------(4)

Similarly Dividing (2)/(3), we get

[n!/(n-r)!r!]/[n!/(n-r-1)!(r+1)!] = 220/66

⇒r + 1/(n - r ) = 10/3

⇒3r + 3 = 10n - 10r

⇒10n - 13r = 3------(5)

Solving (4) and (5), we get

n = 12 and r  = 9.

To find rC₄ = ⁹C₄ = 9!/5!4! = 126

Hope, it helps !