If `""^(n-1)C_(r ): ""^(n)C_(r ): ""^(n+1)C_(r )=6:9:13`, find ` 'n' ` and ` 'r'`.

Answer

To solve the problem where the ratios of combinations are given as \( \binom{n-1}{r} : \binom{n}{r} : \binom{n+1}{r} = 6 : 9 : 13 \), we will approach it step by step.

### Step 1: Set up the equations based on the ratios
From the given ratios, we can express the first two ratios:
\[ \frac{\binom{n-1}{r}}{\binom{n}{r}} = \frac{6}{9} \quad \text{and} \quad \frac{\binom{n}{r}}{\binom{n+1}{r}} = \frac{9}{13} \]

### Step 2: Simplify the first ratio
Using the formula for combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
we can write:
\[ \frac{\binom{n-1}{r}}{\binom{n}{r}} = \frac{\frac{(n-1)!}{r!(n-1-r)!}}{\frac{n!}{r!(n-r)!}} = \frac{(n-1)! \cdot (n-r)!}{n! \cdot (n-1-r)!} \]
This simplifies to:
\[ \frac{(n-1)! \cdot (n-r)!}{n \cdot (n-1)! \cdot (n-1-r)!} = \frac{(n-r)}{n} \]
Setting this equal to \( \frac{6}{9} \):
\[ \frac{n-r}{n} = \frac{2}{3} \]

### Step 3: Cross-multiply and solve for \( n \) and \( r \)
Cross-multiplying gives:
\[ 3(n - r) = 2n \implies 3n - 3r = 2n \implies n = 3r \]
(Equation 1)

### Step 4: Simplify the second ratio
Now we simplify the second ratio:
\[ \frac{\binom{n}{r}}{\binom{n+1}{r}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{(n+1)!}{r!(n+1-r)!}} = \frac{n! \cdot (n+1-r)!}{(n+1)! \cdot (n-r)!} \]
This simplifies to:
\[ \frac{(n+1-r)}{(n+1)} \]
Setting this equal to \( \frac{9}{13} \):
\[ \frac{n+1-r}{n+1} = \frac{9}{13} \]

### Step 5: Cross-multiply and solve for \( n \) and \( r \)
Cross-multiplying gives:
\[ 13(n + 1 - r) = 9(n + 1) \implies 13n + 13 - 13r = 9n + 9 \]
Rearranging gives:
\[ 4n - 13r = -4 \]
(Equation 2)

### Step 6: Solve the system of equations
Now we have two equations:
1. \( n = 3r \)
2. \( 4n - 13r = -4 \)

Substituting \( n = 3r \) into Equation 2:
\[ 4(3r) - 13r = -4 \implies 12r - 13r = -4 \implies -r = -4 \implies r = 4 \]

### Step 7: Find \( n \)
Now substituting \( r = 4 \) back into Equation 1:
\[ n = 3 \cdot 4 = 12 \]

### Final Answer
Thus, the values of \( n \) and \( r \) are:
\[ n = 12, \quad r = 4 \]