Question

n-1Cr+nCr+n+1Cr=6:9:13 find n and r

Answer 1

Answer:

    n = 12, r = 4

Step-by-step explanation:

Given,

    $\mathsf{^{n-1}C_{r}:^{n}C_{r}:^{n+1}C_{r}=6:9:13}$

Taking the first two ratios, we get

    $\mathsf{^{n-1}C_{r}:^{n}C_{r}=6:9}$

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

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

$\mathsf{or,\:\frac{n-r}{n}=\frac{2}{3}}$

or, 3n - 3r = 2n

or, n = 3r ..... (1)

Again taking the last two ratios, we get

    $\mathsf{^{n}C_{r}:^{n+1}C_{r}=9:13}$

$\mathsf{or,\:\frac{n!}{r!(n-r)!}:\frac{(n+1)!}{r!(n+1-r)!}=9:13}$

$\mathsf{or,\:\frac{n!}{r!(n-r)!}\times \frac{r!(n+1-r)(n-r)!}{(n+1)n!}=\frac{9}{13}}$

$\mathsf{or,\:\frac{n+1-r}{n+1}=\frac{9}{13}}$

or, 13 (n + 1 - r) = 9 (n + 1)

or, 13 (3r + 1 - r) = 9 (3r + 1), by (1)

or, 26r + 13 = 27r + 9

or, 27r - 26r = 13 - 9

or, r = 4

From (1), we get

    n = 3 × 4 = 12

Therefore n = 12 and r = 4