1. **State the problem:** We need to find how many different committees of 3 men and 4 women can be formed from 8 men and 6 women. 2. **Formula used:** The number of ways to choose $k$ items from $n$ items is given by the combination formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ 3. **Calculate the number of ways to choose 3 men from 8 men:** $$\binom{8}{3} = \frac{8!}{3!\times 5!}$$ 4. **Calculate the number of ways to choose 4 women from 6 women:** $$\binom{6}{4} = \frac{6!}{4!\times 2!}$$ 5. **Calculate each combination:** $$\binom{8}{3} = \frac{8 \times 7 \times 6 \times \cancel{5!}}{3 \times 2 \times 1 \times \cancel{5!}} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$ $$\binom{6}{4} = \frac{6 \times 5 \times \cancel{4!}}{4 \times 3 \times \cancel{2!}} = \frac{6 \times 5}{2 \times 1} = 15$$ 6. **Multiply the two results to get total number of committees:** $$56 \times 15 = 840$$ **Final answer:** There are **840** different committees of 3 men and 4 women that can be formed from 8 men and 6 women.