If `.^(n+1)C_(r+1.): ^nC_r: ^(n-1)C_(r-1)=11:6:3` find the values of n and r.

Answer

To solve the problem, we need to find the values of \( n \) and \( r \) given the ratio of combinations:

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

### Step 1: Set Up the Ratios

From the given ratios, we can express the first two ratios as:

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

And the second ratio as:

\[ \frac{{^nC_r}}{{^{n-1}C_{r-1}}} = \frac{6}{3} = 2 \]

### Step 2: Use the Combination Formula

Using the combination formula \( ^nC_r = \frac{n!}{r!(n-r)!} \), we can rewrite the first ratio:

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

Substituting these into the first ratio:

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

This simplifies to:

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

### Step 3: Simplify the First Ratio

We can simplify \( \frac{(n+1)!}{n!} = n+1 \):

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

Cross-multiplying gives us:

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

Expanding this:

\[ 6n + 6 = 11r + 11 \]

Rearranging gives us:

\[ 6n - 11r = 5 \quad \text{(Equation 1)} \]

### Step 4: Use the Second Ratio

Now, we analyze the second ratio:

\[ \frac{{^nC_r}}{{^{n-1}C_{r-1}}} = 2 \]

Using the combination formula again:

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

Substituting gives us:

\[ \frac{{\frac{n!}{r!(n-r)!}}}{{\frac{(n-1)!}{(r-1)!(n-r)!}}} = 2 \]

This simplifies to:

\[ \frac{n!}{r!} \cdot \frac{(r-1)!}{(n-1)!} = 2 \]

Thus, we have:

\[ \frac{n}{r} = 2 \]

From this, we can express \( n \) in terms of \( r \):

\[ n = 2r \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1

Now we substitute \( n = 2r \) into Equation 1:

\[ 6(2r) - 11r = 5 \]

This simplifies to:

\[ 12r - 11r = 5 \]

Thus:

\[ r = 5 \]

### Step 6: Find \( n \)

Now substituting \( r = 5 \) back into Equation 2:

\[ n = 2(5) = 10 \]

### Final Answer

The values of \( n \) and \( r \) are:

\[ n = 10, \quad r = 5 \]