To solve the problem, we will use the given ratios of combinations and manipulate them step by step to find the value of \( P(n, r) : C(n, r) \).
### Step 1: Write down the given ratios
We have two ratios given:
1. \( C(n, r) : C(n, r + 1) = 1 : 2 \)
2. \( C(n, r + 1) : C(n, r + 2) = 2 : 3 \)
### Step 2: Express the ratios in terms of combinations
From the first ratio:
\[
\frac{C(n, r)}{C(n, r + 1)} = \frac{1}{2}
\]
This implies:
\[
C(n, r) = \frac{1}{2} C(n, r + 1)
\]
From the second ratio:
\[
\frac{C(n, r + 1)}{C(n, r + 2)} = \frac{2}{3}
\]
This implies:
\[
C(n, r + 1) = \frac{2}{3} C(n, r + 2)
\]
### Step 3: Write the combinations in terms of factorials
Using the formula for combinations:
\[
C(n, r) = \frac{n!}{r!(n-r)!}
\]
We can express \( C(n, r + 1) \) and \( C(n, r + 2) \) similarly:
\[
C(n, r + 1) = \frac{n!}{(r+1)!(n-r-1)!}
\]
\[
C(n, r + 2) = \frac{n!}{(r+2)!(n-r-2)!}
\]
### Step 4: Substitute the combinations into the ratios
From the first ratio:
\[
\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}} = \frac{1}{2}
\]
This simplifies to:
\[
\frac{(r+1)! (n-r-1)!}{r! (n-r)!} = \frac{1}{2}
\]
Cancelling out \( n! \) gives:
\[
\frac{(r+1)(n-r-1)!}{(n-r)(n-r-1)!} = \frac{1}{2}
\]
Thus:
\[
\frac{r+1}{n-r} = \frac{1}{2}
\]
Cross-multiplying gives:
\[
2(r + 1) = n - r
\]
This simplifies to:
\[
3r + 2 = n \quad \text{(Equation 1)}
\]
From the second ratio:
\[
\frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{(r+2)!(n-r-2)!}} = \frac{2}{3}
\]
This simplifies to:
\[
\frac{(r+2)!(n-r-2)!}{(r+1)!(n-r-1)!} = \frac{2}{3}
\]
Cancelling out \( n! \) gives:
\[
\frac{(r+2)(n-r-2)!}{(n-r-1)(n-r-2)!} = \frac{2}{3}
\]
Thus:
\[
\frac{r+2}{n-r-1} = \frac{2}{3}
\]
Cross-multiplying gives:
\[
3(r + 2) = 2(n - r - 1)
\]
Expanding this gives:
\[
3r + 6 = 2n - 2r - 2
\]
Rearranging gives:
\[
5r + 8 = 2n \quad \text{(Equation 2)}
\]
### Step 5: Solve the equations
Now we have two equations:
1. \( n = 3r + 2 \)
2. \( 2n = 5r + 8 \)
Substituting Equation 1 into Equation 2:
\[
2(3r + 2) = 5r + 8
\]
This simplifies to:
\[
6r + 4 = 5r + 8
\]
Rearranging gives:
\[
r = 4
\]
### Step 6: Find \( n \)
Using \( r = 4 \) in Equation 1:
\[
n = 3(4) + 2 = 12 + 2 = 14
\]
### Step 7: Calculate \( P(n, r) : C(n, r) \)
Now we need to find \( P(n, r) \) and \( C(n, r) \):
\[
P(n, r) = \frac{n!}{(n-r)!} = \frac{14!}{(14-4)!} = \frac{14!}{10!} = 14 \times 13 \times 12 \times 11
\]
Calculating this gives:
\[
P(n, r) = 14 \times 13 \times 12 \times 11 = 24024
\]
Now calculate \( C(n, r) \):
\[
C(n, r) = \frac{n!}{r!(n-r)!} = \frac{14!}{4! \times 10!} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001
\]
### Step 8: Find the ratio \( P(n, r) : C(n, r) \)
Now we find:
\[
P(n, r) : C(n, r) = 24024 : 1001
\]
### Final Answer
Thus, the final answer is:
\[
P(n, r) : C(n, r) = 24024 : 1001
\]
