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 :

<p>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 $2n + 5r$.</p> <h3>Step-by-Step Solution:</h3> <ol> <li>Write the given proportions involving binomial coefficients:</li> </ol> <p>$ \frac{n+1}{r+1} \times \binom{n}{r} : \binom{n}{r} : \frac{r}{n} \times \binom{n}{r} = 55 : 35 : 21 $</p> <ol> <li>Simplify the proportions:</li> </ol> <p>$ \frac{n+1}{r+1} = \frac{55}{35} \quad \text{and} \quad \frac{n}{r} = \frac{35}{21} $</p> <ol> <li>From the simplified ratios, we establish the following two equations:</li> </ol> <p>$ \frac{n+1}{r+1} = \frac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad \text{.... (1)} $</p> <p>$ \frac{n}{r} = \frac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad \text{.... (2)} $</p> <ol> <li>Solve equations (1) and (2) simultaneously:</li> </ol> <ul> <li>From equation (2), solve for $n$:</li> </ul> <p>$ 3n = 5r \quad \Rightarrow \quad n = \frac{5r}{3} $</p> <ul> <li>Substitute $n$ into equation (1):</li> </ul> <p>$ 7 \left(\frac{5r}{3}\right) - 11r = 4 \quad \Rightarrow \quad \frac{35r}{3} - 11r = 4 $</p> <p>$ \frac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \frac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6 $</p> <ul> <li>Substituting $r$ back to find $n$:</li> </ul> <p>$ n = \frac{5 \times 6}{3} = 10 $</p> <ol> <li>Compute $2n + 5r$:</li> </ol> <p>$ 2n + 5r = 2 \times 10 + 5 \times 6 = 20 + 30 = 50 $</p> <p>Thus, the value of $2n + 5r$ is 50.</p>