`""^(20)C_(r) =""^(20)C_(r-10),` then find the value of `""^(18)C_(r)`

Answer

To solve the problem \(^{20}C_r = ^{20}C_{r-10}\) and find the value of \(^{18}C_r\), we can follow these steps:

### Step 1: Understand the equation
The equation \(^{20}C_r = ^{20}C_{r-10}\) implies that the number of combinations of choosing \(r\) items from 20 is equal to the number of combinations of choosing \(r-10\) items from 20.

### Step 2: Use the property of combinations
We know that \(^{n}C_k = ^{n}C_{n-k}\). Therefore, we can rewrite the equation as:
\[ ^{20}C_r = ^{20}C_{20 - (r - 10)} \]
This simplifies to:
\[ ^{20}C_r = ^{20}C_{30 - r} \]

### Step 3: Set up the equation
From the equality \(^{20}C_r = ^{20}C_{30 - r}\), we can conclude that:
\[ r = 30 - r \quad \text{or} \quad r = 20 - r + 10 \]
This leads us to:
\[ 2r = 30 \quad \Rightarrow \quad r = 15 \]

### Step 4: Find \(^{18}C_r\)
Now that we have found \(r = 15\), we need to calculate \(^{18}C_{15}\).

### Step 5: Calculate \(^{18}C_{15}\)
Using the formula for combinations:
\[ ^{n}C_k = \frac{n!}{k!(n-k)!} \]
we can substitute \(n = 18\) and \(k = 15\):
\[ ^{18}C_{15} = \frac{18!}{15! \cdot (18 - 15)!} = \frac{18!}{15! \cdot 3!} \]

### Step 6: Simplify the factorials
Now, we can simplify:
\[ ^{18}C_{15} = \frac{18 \times 17 \times 16 \times 15!}{15! \times 3!} \]
The \(15!\) cancels out:
\[ ^{18}C_{15} = \frac{18 \times 17 \times 16}{3!} \]
Calculating \(3!\):
\[ 3! = 6 \]
So,
\[ ^{18}C_{15} = \frac{18 \times 17 \times 16}{6} \]

### Step 7: Perform the multiplication and division
Calculating the numerator:
\[ 18 \times 17 = 306 \]
\[ 306 \times 16 = 4896 \]
Now divide by 6:
\[ ^{18}C_{15} = \frac{4896}{6} = 816 \]

### Final Answer
Thus, the value of \(^{18}C_{15}\) is \(816\).

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