Question

Let $0 \leq r \leq n$. If $^nC_{r+1} : ^nC_r : ^nC_{r-1} = 55 : 35 : 21,$then $2n + 5r$ is equal to:

• A. 60
• B. 62
• C. 50
• D. 55

Answer 1

Answer: (C)

50

Answer 2

$\frac{{n+1}C_r}{nC_r} = \frac{55}{35}$

$\frac{(n+1)!}{(r+1)!(n-r)!} \div \frac{n!}{r!(n-r)!} = \frac{11}{7}$

$\frac{n+1}{r+1} = \frac{11}{7}$

$7n = 4 + 11r$

$\frac{nC_r}{n-1C_{r-1}} = \frac{35}{21}$

$\frac{n!}{r!(n-r)!} \div \frac{(n-1)!}{(r-1)!(n-r)!} = \frac{5}{3}$

$\frac{n}{r} = \frac{5}{3}$

$3n = 5r$

Solving for $r = 6$ and $n = 10$:

$2n + 5r = 50$