Let [0 r n].

Question

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

Accepted Answer

Given $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then we are to determine the value of $2n + 5r$.

Write the given proportions involving binomial coefficients:

$    \frac{n+1}{r+1} 	imes \binom{n}{r} : \binom{n}{r} : \frac{r}{n} 	imes \binom{n}{r} = 55 : 35 : 21    $

Simplify the proportions:

$    \frac{n+1}{r+1} = \frac{55}{35} \quad 	ext{and} \quad \frac{n}{r} = \frac{35}{21}    $

From the simplified ratios, we establish the following two equations:

$    \frac{n+1}{r+1} = \frac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad 	ext{.... (1)}    $

$    \frac{n}{r} = \frac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad 	ext{.... (2)}    $

Solve equations (1) and (2) simultaneously:

From equation (2), solve for $n$:

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

Substitute $n$ into equation (1):

$      7 \left(\frac{5r}{3}ight) - 11r = 4 \quad \Rightarrow \quad \frac{35r}{3} - 11r = 4      $

$      \frac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \frac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6      $

Substituting $r$ back to find $n$:

$      n = \frac{5 	imes 6}{3} = 10      $

Compute $2n + 5r$:

$    2n + 5r = 2 	imes 10 + 5 	imes 6 = 20 + 30 = 50    $

Thus, the value of $2n + 5r$ is 50.

Solutions · 1