A committee of 6 is to be formed from 6 men and 5 women. If the committee must have at least 2 women, and a particular man and a particular woman refuse to serve together, how many committees are possible?

Step 1: Setup

We have:

• 6 men (M)

• 5 women (W)

• Committee size = 6

• Condition 1: At least 2 women

• Condition 2: A particular man (say M∗) and a particular woman (say W∗) refuse to serve together.

🔵 Step 2: Total Committees with ≥2 Women (ignoring restriction)

We count committees with at least 2 women:

Total=∑k=25(5k)⋅(66−k)

• k=2: (52)(64)=10⋅15=150

• k=3: (53)(63)=10⋅20=200

• k=4: (54)(62)=5⋅15=75

• k=5: (55)(61)=1⋅6=6

Total=150+200+75+6=431

So without restriction, 431 committees.

🔴 Step 3: Subtract Invalid Committees (where M∗ and W∗ both included)

If both M∗ and W∗ are chosen, then:

• We already have 1 man and 1 woman fixed.

• Remaining: choose 4 from the other 9 people (5 men left, 4 women left).

• Condition: total women ≥ 2 → since W∗ is already included, we need at least 1 more woman among the 4 chosen.

So count = total committees with both fixed − committees with no additional women.

• Total with both fixed: (94)=126

• No additional women (all 4 chosen are men): (54)=5

So invalid = 126−5=121.

🟣 Step 4: Final Answer

431−121=310

✅ There are 310 possible committees.