Find the number of permutations of n things taken r at a time in which two particular things :
(i) always occur
(ii) never occur.

Answer

To solve the problem of finding the number of permutations of \( n \) things taken \( r \) at a time in which two particular things either always occur or never occur, we can break it down into two parts.

### Part (i): Two Particular Things Always Occur

1. **Understanding the Problem**: We have \( n \) items, and we want to find permutations of \( r \) items where two specific items must always be included.

2. **Fix the Two Particular Things**: Since the two particular items must always occur, we can treat them as a single entity or block. This means we now have \( n - 1 \) items to choose from (the block of two items counts as one).

3. **Calculate the Remaining Items**: We need to select \( r - 2 \) more items from the remaining \( n - 2 \) items (since we have already included the two particular items).

4. **Permutations of the Selected Items**: The total number of items we are arranging now is \( r - 1 \) (the block of two items plus \( r - 2 \) other items). The number of ways to arrange these \( r - 1 \) items is given by \( (r - 1)! \).

5. **Arrangements of the Two Particular Items**: The two particular items can be arranged among themselves in \( 2! \) ways.

6. **Final Calculation**: Therefore, the total number of permutations where the two particular items always occur is given by:
\[ P(n, r) = (n - 2)P(r - 2) \times 2! \times (r - 1)! \]

### Part (ii): Two Particular Things Never Occur

1. **Understanding the Problem**: Now, we want to find permutations of \( r \) items where the two particular items are not included.

2. **Total Permutations Without Restrictions**: The total permutations of \( n \) items taken \( r \) at a time is given by \( P(n, r) \).

3. **Permutations with the Two Particular Items**: The number of permutations where the two particular items are included (as calculated in part (i)) is:
\[ P(n - 2, r - 2) \times 2! \times (r - 1)! \]

4. **Calculate Permutations Where They Never Occur**: To find the permutations where the two particular items never occur, we subtract the number of permutations where they do occur from the total permutations:
\[ P(n, r) - \left( P(n - 2, r - 2) \times 2! \times (r - 1)! \right) \]

5. **Final Calculation**: Therefore, the total number of permutations where the two particular items never occur is:
\[ P(n, r) - \left( (n - 2)P(r - 2) \times 2! \times (r - 1)! \right) \]

### Summary of Results
- **Part (i)**: The number of permutations where two particular items always occur:
\[ P(n, r) = (n - 2)P(r - 2) \times 2! \times (r - 1)! \]
- **Part (ii)**: The number of permutations where two particular items never occur:
\[ P(n, r) - \left( (n - 2)P(r - 2) \times 2! \times (r - 1)! \right) \]