Question

Let \( 0 \le r \le n \). If \( ^{n+1}C_{r+1} : ^nC_r : ^{n-1}C_{r-1} = 55 : 35 : 21 \), then \( 2n + 5r \) is equal to:

• A. 60
• B. 62
• C. 50
• D. 55

Answer 1

The Correct Option is C

We are given the ratio of three binomial coefficients, \(^{n+1}C_{r+1} : ^nC_r : ^{n-1}C_{r-1} = 55 : 35 : 21\), for \(0 \le r \le n\). The objective is to find the value of the expression \(2n + 5r\).

Concept Used:

The solution uses the property of the ratio of two binomial coefficients. While the standard formula \(\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}\) is useful, in this problem, the upper and lower indices change simultaneously. Therefore, it is more direct to use the factorial definition of a binomial coefficient, \(^nC_r = \frac{n!}{r!(n-r)!}\), to derive the required ratios.

The key ratios we will use are:

1. \( \frac{^nC_r}{^{n-1}C_{r-1}} = \frac{n}{r} \)
2. \( \frac{^{n+1}C_{r+1}}{^nC_r} = \frac{n+1}{r+1} \)

Step-by-Step Solution:

Step 1: Set up two separate equations from the given compound ratio.

From the ratio \(^nC_r : ^{n-1}C_{r-1} = 35 : 21\), we can write the first equation:

\[ \frac{^nC_r}{^{n-1}C_{r-1}} = \frac{35}{21} = \frac{5}{3} \]

From the ratio \(^{n+1}C_{r+1} : ^nC_r = 55 : 35\), we can write the second equation:

\[ \frac{^{n+1}C_{r+1}}{^nC_r} = \frac{55}{35} = \frac{11}{7} \]

Step 2: Simplify the first ratio using the factorial definition to find a relation between \(n\) and \(r\).

\[ \frac{^nC_r}{^{n-1}C_{r-1}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{(n-1)!}{(r-1)!(n-1-(r-1))!}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{(n-1)!}{(r-1)!(n-r)!}} \] \[ = \frac{n!}{r!(n-r)!} \times \frac{(r-1)!(n-r)!}{(n-1)!} = \frac{n \cdot (n-1)!}{r \cdot (r-1)!} \times \frac{(r-1)!}{(n-1)!} = \frac{n}{r} \]

Equating this result to the value from Step 1:

\[ \frac{n}{r} = \frac{5}{3} \implies 3n = 5r \quad \text{(Equation 1)} \]

Step 3: Simplify the second ratio to find another relation between \(n\) and \(r\).

\[ \frac{^{n+1}C_{r+1}}{^nC_r} = \frac{\frac{(n+1)!}{(r+1)!(n+1-(r+1))!}}{\frac{n!}{r!(n-r)!}} = \frac{\frac{(n+1)!}{(r+1)!(n-r)!}}{\frac{n!}{r!(n-r)!}} \] \[ = \frac{(n+1)!}{(r+1)!(n-r)!} \times \frac{r!(n-r)!}{n!} = \frac{(n+1) \cdot n!}{(r+1) \cdot r!} \times \frac{r!}{n!} = \frac{n+1}{r+1} \]

Equating this result to the value from Step 1:

\[ \frac{n+1}{r+1} = \frac{11}{7} \implies 7(n+1) = 11(r+1) \] \[ 7n + 7 = 11r + 11 \implies 7n - 11r = 4 \quad \text{(Equation 2)} \]

Step 4: Solve the system of linear equations for \(n\) and \(r\).

From Equation 1, we have \(n = \frac{5}{3}r\). Substitute this into Equation 2:

\[ 7\left(\frac{5}{3}r\right) - 11r = 4 \] \[ \frac{35r}{3} - 11r = 4 \]

Multiplying by 3 to clear the denominator:

\[ 35r - 33r = 12 \] \[ 2r = 12 \implies r = 6 \]

Now substitute \(r=6\) back into the expression for \(n\):

\[ n = \frac{5}{3}(6) = 10 \]

The values are \(n=10\) and \(r=6\). These satisfy the condition \(0 \le r \le n\).

Step 5: Calculate the value of the expression \(2n + 5r\).

Substitute the obtained values of \(n\) and \(r\):

\[ 2n + 5r = 2(10) + 5(6) \] \[ = 20 + 30 = 50 \]

Thus, the value of \(2n + 5r\) is 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 \( n = 10 \):

\[ 2n + 5r = 50 \]