(i) `""^(50)C_(47)=` ……….
(ii) `""^(15)C_(14)=` ………….

Answer

To solve the given problems, we will use the formula for combinations, which is:

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

### Part (i): Calculate \( ^{50}C_{47} \)

1. **Identify n and r**: Here, \( n = 50 \) and \( r = 47 \).
2. **Use the formula**:
\[ ^{50}C_{47} = \frac{50!}{47!(50-47)!} = \frac{50!}{47! \cdot 3!} \]
3. **Simplify the factorials**:
\[ 50! = 50 \times 49 \times 48 \times 47! \]
Therefore, we can write:
\[ ^{50}C_{47} = \frac{50 \times 49 \times 48 \times 47!}{47! \cdot 3!} \]
4. **Cancel out \( 47! \)**:
\[ ^{50}C_{47} = \frac{50 \times 49 \times 48}{3!} \]
5. **Calculate \( 3! \)**:
\[ 3! = 3 \times 2 \times 1 = 6 \]
6. **Substitute back**:
\[ ^{50}C_{47} = \frac{50 \times 49 \times 48}{6} \]
7. **Perform the multiplication**:
\[ 50 \times 49 = 2450 \]
\[ 2450 \times 48 = 117600 \]
8. **Divide by 6**:
\[ ^{50}C_{47} = \frac{117600}{6} = 19600 \]

### Part (ii): Calculate \( ^{15}C_{14} \)

1. **Identify n and r**: Here, \( n = 15 \) and \( r = 14 \).
2. **Use the formula**:
\[ ^{15}C_{14} = \frac{15!}{14!(15-14)!} = \frac{15!}{14! \cdot 1!} \]
3. **Simplify the factorials**:
\[ 15! = 15 \times 14! \]
Therefore, we can write:
\[ ^{15}C_{14} = \frac{15 \times 14!}{14! \cdot 1!} \]
4. **Cancel out \( 14! \)**:
\[ ^{15}C_{14} = \frac{15}{1!} \]
5. **Calculate \( 1! \)**:
\[ 1! = 1 \]
6. **Substitute back**:
\[ ^{15}C_{14} = \frac{15}{1} = 15 \]

### Final Answers:
- \( ^{50}C_{47} = 19600 \)
- \( ^{15}C_{14} = 15 \)