`""^(25)C_(22)- ""^(24)C_(21)=` …………

Answer

To solve the expression \( \binom{25}{22} - \binom{24}{21} \), we will use the formula for combinations, which is given by:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

### Step-by-Step Solution:

1. **Calculate \( \binom{25}{22} \)**:
\[ \binom{25}{22} = \frac{25!}{22!(25-22)!} = \frac{25!}{22! \cdot 3!} \]
Here, \( 25 - 22 = 3 \).

2. **Simplify \( \binom{25}{22} \)**:
\[ \binom{25}{22} = \frac{25 \times 24 \times 23 \times 22!}{22! \cdot 3!} \]
The \( 22! \) in the numerator and denominator cancels out:
\[ = \frac{25 \times 24 \times 23}{3!} \]
Since \( 3! = 6 \):
\[ = \frac{25 \times 24 \times 23}{6} \]

3. **Calculate \( \binom{24}{21} \)**:
\[ \binom{24}{21} = \frac{24!}{21!(24-21)!} = \frac{24!}{21! \cdot 3!} \]
Here, \( 24 - 21 = 3 \).

4. **Simplify \( \binom{24}{21} \)**:
\[ \binom{24}{21} = \frac{24 \times 23 \times 22 \times 21!}{21! \cdot 3!} \]
The \( 21! \) in the numerator and denominator cancels out:
\[ = \frac{24 \times 23 \times 22}{3!} \]
Since \( 3! = 6 \):
\[ = \frac{24 \times 23 \times 22}{6} \]

5. **Now, substitute back into the original expression**:
\[ \binom{25}{22} - \binom{24}{21} = \frac{25 \times 24 \times 23}{6} - \frac{24 \times 23 \times 22}{6} \]

6. **Factor out the common terms**:
\[ = \frac{1}{6} \left( 25 \times 24 \times 23 - 24 \times 23 \times 22 \right) \]
\[ = \frac{1}{6} \times 24 \times 23 \times (25 - 22) \]
\[ = \frac{1}{6} \times 24 \times 23 \times 3 \]

7. **Calculate the final result**:
\[ = \frac{1}{6} \times 72 \times 23 \]
\[ = 12 \times 23 = 276 \]

### Final Answer:
\[ \binom{25}{22} - \binom{24}{21} = 276 \]