A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 woman. In how many different ways can it be done?

### Concept & Complementary Counting When tasked with finding combinations with "at least 1" item from a subgroup, calculate the total unrestricted combinations and subtract the combinations where the subgroup is completely excluded. ### Step-by-Step Solution 1. **Total Unrestricted Groups:** We are selecting 4 people from a total pool of 14 people (8 men + 6 women). $$ ^{14}C_4 = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001 \text{ ways} $$ 2. **Groups with NO Women:** Select 4 people entirely from the 8 men. $$ ^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \text{ ways} $$ 3. **Groups with At Least 1 Woman:** Subtract the "no women" scenario from the total. $$ 1001 - 70 = 931 \text{ ways} $$ ### Exam Strategy & Shortcut Notice the denominators when expanding combinations: $4 \times 3 \times 2 \times 1 = 24$. In $^{14}C_4$, $12$ in the numerator cancels beautifully with $24$ to leave $2$, which further cancels with $14$ to leave $7$. Factoring out rather than multiplying large numbers prevents arithmetic mistakes. ### Common Pitfall Failing to combine the total pool correctly (8+6=14) or incorrectly applying permutation formula instead of combination. ### Final Answer Therefore, the correct answer is **931**.