Evaluate
(i) `""^(10)C_(4)+""^(10)C_(5)`
(ii) `""^(13)C_(6)+ ""^(13)C_(5)`
(iii) `""^(19)C_(17)+ ""^(19)C_(18)`.

Answer

To evaluate the given expressions, we will use the identity:

\[ nC_r + nC_{r+1} = (n+1)C_{r+1} \]

### (i) Evaluate \(10C_4 + 10C_5\)

1. **Identify \(n\) and \(r\)**:
- Here, \(n = 10\) and \(r = 4\).

2. **Apply the identity**:
\[ 10C_4 + 10C_5 = 11C_5 \]

3. **Calculate \(11C_5\)**:
\[ 11C_5 = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!} \]

4. **Simplify**:
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{5! \times 6!} \]
The \(6!\) cancels out:
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{5!} \]
Where \(5! = 120\).

5. **Calculate**:
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{120} \]
- Calculate \(11 \times 10 = 110\)
- \(110 \times 9 = 990\)
- \(990 \times 8 = 7920\)
- \(7920 \times 7 = 55440\)

6. **Final Division**:
\[ \frac{55440}{120} = 462 \]

### Answer for (i):
\[ 10C_4 + 10C_5 = 462 \]

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### (ii) Evaluate \(13C_6 + 13C_5\)

1. **Identify \(n\) and \(r\)**:
- Here, \(n = 13\) and \(r = 5\).

2. **Apply the identity**:
\[ 13C_6 + 13C_5 = 14C_6 \]

3. **Calculate \(14C_6\)**:
\[ 14C_6 = \frac{14!}{6!(14-6)!} = \frac{14!}{6! \cdot 8!} \]

4. **Simplify**:
\[ = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8!}{6! \times 8!} \]
The \(8!\) cancels out:
\[ = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9}{6!} \]
Where \(6! = 720\).

5. **Calculate**:
\[ = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9}{720} \]
- Calculate \(14 \times 13 = 182\)
- \(182 \times 12 = 2184\)
- \(2184 \times 11 = 24024\)
- \(24024 \times 10 = 240240\)
- \(240240 \times 9 = 2162160\)

6. **Final Division**:
\[ \frac{2162160}{720} = 3003 \]

### Answer for (ii):
\[ 13C_6 + 13C_5 = 3003 \]

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### (iii) Evaluate \(19C_{17} + 19C_{18}\)

1. **Identify \(n\) and \(r\)**:
- Here, \(n = 19\) and \(r = 17\).

2. **Apply the identity**:
\[ 19C_{17} + 19C_{18} = 20C_{18} \]

3. **Calculate \(20C_{18}\)**:
\[ 20C_{18} = \frac{20!}{18!(20-18)!} = \frac{20!}{18! \cdot 2!} \]

4. **Simplify**:
\[ = \frac{20 \times 19 \times 18!}{18! \cdot 2!} \]
The \(18!\) cancels out:
\[ = \frac{20 \times 19}{2!} = \frac{20 \times 19}{2} = \frac{380}{2} = 190 \]

### Answer for (iii):
\[ 19C_{17} + 19C_{18} = 190 \]

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### Summary of Answers:
1. \(10C_4 + 10C_5 = 462\)
2. \(13C_6 + 13C_5 = 3003\)
3. \(19C_{17} + 19C_{18} = 190\)

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