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

The Correct Option is C

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

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

\[ \frac{n+1}{r+1} = \frac{11}{7} \]

\[ 7n = 4 + 11r \]

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

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

\[ \frac{n}{r} = \frac{5}{3} \]

\[ 3n = 5r \]

Solving for \( r = 6 \) and \( n = 10 \):

\[ 2n + 5r = 50 \]