The number of ways is 30.

Interpreting the question

The standard version of this question is:

A group has 5 men and 5 women. In how many ways can we select 3 men and 2 women if one particular man and one particular woman must always be included?

Under that interpretation, the answer is 30.

Step-by-step reasoning

  1. One specific man (say M1M_1M1​) and one specific woman (say W1W_1W1​) must always be in the group.
  2. We need a total of 3 men and 2 women.
    • Men: We already have M1M_1M1​, so we must choose 2 more men from the remaining 4 men.
    • Women: We already have W1W_1W1​, so we must choose 1 more woman from the remaining 5 women.
  3. Number of ways to choose the remaining men:

(42)=6\binom{4}{2}=6(24​)=6

  1. Number of ways to choose the remaining woman:

(51)=5\binom{5}{1}=5(15​)=5

  1. Total number of ways (multiply the choices):

6×5=306\times 5=306×5=30

So there are 30 possible committees of 3 men and 2 women with that particular man and woman always included.

HTML table of the counting

Here is a compact HTML table summarizing the selection:

html

<table>
  <tr>
    <th>Category</th>
    <th>Already fixed</th>
    <th>Still to choose</th>
    <th>Choices</th>
    <th>Number of ways</th>
  </tr>
  <tr>
    <td>Men</td>
    <td>1 fixed man</td>
    <td>2 more men from remaining 4</td>
    <td>C(4, 2)</td>
    <td>6</td>
  </tr>
  <tr>
    <td>Women</td>
    <td>1 fixed woman</td>
    <td>1 more woman from remaining 5</td>
    <td>C(5, 1)</td>
    <td>5</td>
  </tr>
  <tr>
    <td colspan="4"><strong>Total ways</strong></td>
    <td><strong>6 × 5 = 30</strong></td>
  </tr>
</table>

Mini “story” to remember the idea

Imagine you’re forming a small event-organizing team: the experienced senior man and senior woman must be on every committee.

  • First, you “lock in” these two.
  • Then you only worry about picking the remaining people:
    • any 2 from the other 4 men,
    • and any 1 from the other 5 women.

Thinking of it this way helps you remember to fix the required people first, then count combinations for the rest.