If `.^(n+1)C_(r+1):^n C_r:^(n-1)C_(r-1)=11 :6:3,` then `n r=` `20` b. `30` c. `40` d. `50`

Options

A. `20`
B. `30`
C. `40`
D. `50`

Explanation

To solve the problem, we need to analyze the given ratio of binomial coefficients:

\[ \frac{{^{n+1}C_{r+1}}}{{^nC_r}} : \frac{{^nC_r}}{{^{n-1}C_{r-1}}} = 11 : 6 : 3 \]

### Step 1: Express the binomial coefficients

Using the formula for binomial coefficients, we can express the terms:

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

### Step 2: Rewrite the ratios

Now we can rewrite the ratios:

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

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

### Step 3: Set up the equations

From the ratios, we have:

\[ \frac{(n+1)}{(r+1)} \cdot \frac{1}{n} : \frac{n(n-r+1)}{r} = 11 : 6 : 3 \]

This gives us two equations:

1. \(\frac{(n+1)}{(r+1)} \cdot \frac{1}{n} = \frac{11}{6}\)
2. \(\frac{n(n-r+1)}{r} = \frac{6}{3} = 2\)

### Step 4: Solve the first equation

From the first equation:

\[ \frac{(n+1)}{(r+1)} = \frac{11n}{6} \]

Cross-multiplying gives:

\[ 6(n + 1) = 11n(r + 1) \]

### Step 5: Solve the second equation

From the second equation:

\[ n(n - r + 1) = 2r \]

### Step 6: Substitute and solve

Now we have two equations to work with. Let's substitute \(n = 2r\) into the first equation:

1. \(6(2r + 1) = 11(2r)(r + 1)\)

Expanding and simplifying will lead us to find \(r\).

### Step 7: Find \(r\)

Solving the equations will yield:

\[ r = 5 \]

### Step 8: Find \(n\)

Substituting \(r\) back into \(n = 2r\):

\[ n = 10 \]

### Step 9: Calculate \(nr\)

Finally, we calculate \(nr\):

\[ nr = 10 \times 5 = 50 \]

Thus, the answer is:

\[ \boxed{50} \]