A box contains 10 pens, 6 pencils, and 8 erasers. In how many ways can you select 3 writing instruments from the box, if at least one pen and one pencil are to be included in the selection?

The problem requires finding how many ways you can select 3 writing instruments from a box containing 10 pens, 6 pencils, and 8 erasers, with the condition that at least one pen and one pencil must be included in the selection.

### Step 1: Total writing instruments
- Pens: 10
- Pencils: 6
- Erasers: 8

So, the total number of writing instruments is:
\[
10 + 6 + 8 = 24
\]

### Step 2: Breakdown of valid cases
We need to select 3 writing instruments where at least one is a pen and at least one is a pencil. There are two possible valid combinations for this:

1. **Case 1**: Select 1 pen, 1 pencil, and 1 more item (which could be a pen, pencil, or eraser).
2. **Case 2**: Select 2 pens and 1 pencil.

#### Case 1: Selecting 1 pen, 1 pencil, and 1 other item
- Select 1 pen: The number of ways to choose 1 pen from 10 is \( \binom{10}{1} = 10 \).
- Select 1 pencil: The number of ways to choose 1 pencil from 6 is \( \binom{6}{1} = 6 \).
- Select 1 other item (pen, pencil, or eraser): The number of ways to choose 1 other item from all the items (24 total minus the 2 already chosen pens and pencils) is:
\[
10 + 6 + 8 = 24 - 2 = 22
\]
So the number of ways to choose the 3rd item from 22 remaining items is 22. Hence, the total number of ways for this case is:
\[
10 \times 6 \times 22 = 1320
\]



Find out how to select 3 writing instruments from a box containing 10 pens, 6 pencils, and 8 erasers, ensuring at least one pen and one pencil are included. This problem explores combinatorics, including combinations without repetition. Suitable for Grades 9-12.

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