Question

If nCr : (n+1)Cr : (n+2)Cr = 1:3:7, find the values of n and r.

Answer 1

The solution is n = 5 and r = 4

Step-by-step explanation:

Given,

nCr : (n+1)Cr : (n+2)Cr = 1:3:7

To find the value of n and r.

Formula

nCr = $\frac{n!}{r!(n-r)!}$  [ n! means = n.(n-1).(n-2)....3.2.1]

Now,

nCr : (n+1)Cr = 1:3                 and (n+1)Cr : (n+2)Cr = 3:7

or, $\frac{n!}{r!(n-r)!}$ : $\frac{(n+1)!}{r!(n+1-r)!}$ = 1:3         or,  $\frac{(n+1)!}{r!(n+1-r)!}$ : $\frac{(n+2)!}{r!(n+2-r)!}$ = $\frac{3}{7}$

or,  $\frac{n!}{r!(n-r)!}$ × $\frac{r!(n+1-r)!}{(n+1)!}$ = $\frac{1}{3}$          or, $\frac{(n+1)!}{r!(n+1-r)!}$ ×$\frac{r!(n+2-r)!}{(n+2)!}$ =  $\frac{3}{7}$  

or, $\frac{n+1-r}{n+1}$ = $\frac{1}{3}$                               or, $\frac{n+2-r}{n+2}$ =  $\frac{3}{7}$

or, 3(n+1-r) = n+1                        or, 7(n+2-r) = 3(n+2)

or, 3n+3-3r = n+1                       or, 7n+14-7r = 3n+6

or, 2n-3r = -2                             or, 4n-7r = -8

Now, by solving

2n-3r = -2 -----(1)

4n-7r = -8 -----(2) we will get n and r

Multiplying (1) by and then subtract with (2) we get,

2(2n-3r) - (4n-7r) = -4-(-8)

or, 4n-6r-4n+7r = 4

or, r = 4

From (1) we get,

2n = -2+3(4)

or, 2n = 10

or, n = 5

Hence,

n = 5 and r = 4