IF `""^8C_r-""^7 C_3= "^7 C_2` then r=

Options

A. 3
B. 4
C. 5
D. 6

Answer

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

### Step 1: Calculate \( \binom{7}{3} \) and \( \binom{7}{2} \)

Using the formula for combinations, \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \):

\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 2: Substitute the values into the equation

Now we substitute \( \binom{7}{3} \) and \( \binom{7}{2} \) into the original equation:

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

### Step 3: Solve for \( \binom{8}{r} \)

Rearranging the equation gives:

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

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

### 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 \( r = 4 \): \( \binom{8}{4} = 70 \)

We find that \( \binom{8}{3} = 56 \).

### Step 5: Conclusion

Thus, the value of \( r \) is:

\[ r = 3 \]