Evaluate: `.^(20)C_(5)+^(20)C_(4)+^(21)C_(4)+^(22)C_(4)`

Answer

To evaluate the expression \( \binom{20}{5} + \binom{20}{4} + \binom{21}{4} + \binom{22}{4} \), we can use properties of combinations.

### Step-by-Step Solution:

1. **Identify the Combinations**:
We have the terms \( \binom{20}{5} \), \( \binom{20}{4} \), \( \binom{21}{4} \), and \( \binom{22}{4} \).

2. **Use the Combination Identity**:
We can use the identity \( \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \).
- First, combine \( \binom{20}{5} \) and \( \binom{20}{4} \):
\[ \binom{20}{5} + \binom{20}{4} = \binom{21}{5} \]

3. **Combine with the Next Terms**:
Now we have:
\[ \binom{21}{5} + \binom{21}{4} \]
Using the same identity again:
\[ \binom{21}{5} + \binom{21}{4} = \binom{22}{5} \]

4. **Combine with the Last Term**:
Now we have:
\[ \binom{22}{5} + \binom{22}{4} \]
Again applying the identity:
\[ \binom{22}{5} + \binom{22}{4} = \binom{23}{5} \]

5. **Calculate \( \binom{23}{5} \)**:
Now we need to calculate \( \binom{23}{5} \):
\[ \binom{23}{5} = \frac{23!}{5!(23-5)!} = \frac{23!}{5! \cdot 18!} \]
This can be simplified as:
\[ \binom{23}{5} = \frac{23 \times 22 \times 21 \times 20 \times 19}{5 \times 4 \times 3 \times 2 \times 1} \]

6. **Perform the Calculation**:
- Calculate the numerator:
\[ 23 \times 22 = 506 \]
\[ 506 \times 21 = 10626 \]
\[ 10626 \times 20 = 212520 \]
\[ 212520 \times 19 = 4037880 \]
- Calculate the denominator:
\[ 5! = 120 \]
- Now divide:
\[ \frac{4037880}{120} = 33649 \]

### Final Answer:
Thus, the value of \( \binom{20}{5} + \binom{20}{4} + \binom{21}{4} + \binom{22}{4} \) is \( 33649 \).