Evaluate `(n!)/((n-r)!)`, when :
(i) `n=6, r=2`
(ii) `n=9, r=5`

Answer

To evaluate the expression \(\frac{n!}{(n-r)!}\) for the given values of \(n\) and \(r\), we will follow these steps:

### Part (i): \(n = 6\), \(r = 2\)

1. **Substitute the values into the formula**:
\[ \frac{n!}{(n-r)!} = \frac{6!}{(6-2)!} = \frac{6!}{4!} \]

2. **Expand the factorials**:
\[ 6! = 6 \times 5 \times 4! \]
So, we can rewrite the expression:
\[ \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} \]

3. **Cancel \(4!\) from the numerator and denominator**:
\[ \frac{6 \times 5 \times 4!}{4!} = 6 \times 5 \]

4. **Calculate the result**:
\[ 6 \times 5 = 30 \]

Thus, the answer for part (i) is **30**.

### Part (ii): \(n = 9\), \(r = 5\)

1. **Substitute the values into the formula**:
\[ \frac{n!}{(n-r)!} = \frac{9!}{(9-5)!} = \frac{9!}{4!} \]

2. **Expand the factorials**:
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4! \]
So, we can rewrite the expression:
\[ \frac{9!}{4!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} \]

3. **Cancel \(4!\) from the numerator and denominator**:
\[ \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} = 9 \times 8 \times 7 \times 6 \times 5 \]

4. **Calculate the result step by step**:
- First, calculate \(6 \times 5 = 30\)
- Next, calculate \(30 \times 7 = 210\)
- Then, calculate \(210 \times 8 = 1680\)
- Finally, calculate \(1680 \times 9 = 15120\)

Thus, the answer for part (ii) is **15120**.

### Summary of Answers:
- Part (i): 30
- Part (ii): 15120