If `\ ^8C_r-\ ^7C_3\ =^7C_2\ ` find `r`

Options

A. 2
B. 3
C. 4
D. None of these

Explanation

To solve the equation \( \binom{8}{r} - \binom{7}{3} = \binom{7}{2} \), we will follow these steps:

### Step 1: Write down the known values
First, we need to calculate the values of \( \binom{7}{3} \) and \( \binom{7}{2} \).

\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 2: Substitute the values into the equation
Now, substitute these values into the original equation:

\[ \binom{8}{r} - 35 = 21 \]

### Step 3: Solve for \( \binom{8}{r} \)
Add 35 to both sides of the equation:

\[ \binom{8}{r} = 21 + 35 \]

\[ \binom{8}{r} = 56 \]

### Step 4: Find \( r \) such that \( \binom{8}{r} = 56 \)
Now we need to find \( r \) such that \( \binom{8}{r} = 56 \).

Calculating \( \binom{8}{r} \) for different values of \( r \):

- For \( r = 0 \): \( \binom{8}{0} = 1 \)
- For \( r = 1 \): \( \binom{8}{1} = 8 \)
- For \( r = 2 \): \( \binom{8}{2} = 28 \)
- For \( r = 3 \): \( \binom{8}{3} = 56 \)
- For \( r = 4 \): \( \binom{8}{4} = 70 \)

From the calculations, we see that \( r = 3 \) satisfies the equation \( \binom{8}{r} = 56 \).

### Step 5: Check for other possible values of \( r \)
Using the property of combinations, \( \binom{n}{r} = \binom{n}{n-r} \):

- For \( r = 5 \): \( \binom{8}{5} = \binom{8}{3} = 56 \)

Thus, the possible values for \( r \) are \( 3 \) and \( 5 \).

### Final Answer
The values of \( r \) are \( r = 3 \) or \( r = 5 \).

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