If ` ""^(n)C_r : ""^(n)C_(r+1) =1 : 2 and^(n)C_(r+1): ""^(n)C_(r+2) = 2 : 3 ` determine the values of n and r.

Answer

To solve the problem, we need to determine the values of \( n \) and \( r \) given the ratios of combinations. We have the following equations based on the problem statement:

1. \( \frac{{^nC_r}}{{^nC_{r+1}}} = \frac{1}{2} \)
2. \( \frac{{^nC_{r+1}}}{{^nC_{r+2}}} = \frac{2}{3} \)

### Step 1: Write the combination formulas

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

1. From the first equation:
\[ \frac{{^nC_r}}{{^nC_{r+1}}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}} = \frac{(r+1)!(n-r-1)!}{r!(n-r)!} = \frac{(n-r)}{(r+1)} = \frac{1}{2} \]

2. From the second equation:
\[ \frac{{^nC_{r+1}}}{{^nC_{r+2}}} = \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{(r+2)!(n-r-2)!}} = \frac{(r+2)!(n-r-2)!}{(r+1)!(n-r-1)!} = \frac{(r+2)}{(n-r-1)} = \frac{2}{3} \]

### Step 2: Set up the equations

From the first equation:
\[ 2(n - r) = r + 1 \quad \text{(Cross-multiplying)} \]
This simplifies to:
\[ 2n - 2r = r + 1 \implies 2n = 3r + 1 \quad \text{(Equation 1)} \]

From the second equation:
\[ 3(r + 2) = 2(n - r - 1) \quad \text{(Cross-multiplying)} \]
This simplifies to:
\[ 3r + 6 = 2n - 2r - 2 \implies 5r - 2n = -8 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations:
1. \( 2n = 3r + 1 \)
2. \( 5r - 2n = -8 \)

Substituting \( 2n \) from Equation 1 into Equation 2:
\[ 5r - (3r + 1) = -8 \]
This simplifies to:
\[ 5r - 3r - 1 = -8 \implies 2r - 1 = -8 \implies 2r = -7 \implies r = 12 \]

Now substitute \( r = 12 \) back into Equation 1 to find \( n \):
\[ 2n = 3(12) + 1 = 36 + 1 = 37 \implies n = \frac{37}{2} = 34 \]

### Final Answer
Thus, the values are:
\[ n = 34, \quad r = 12 \]