How many ways can you arrange 8 people at a round table (clockwise and anticlockwise arrangements are considered different) with no restrictions?

1. How many ways can you form a 4-person committee from 6 men and 5 women?

a) with no restrictions We need to choose 4 people from a total of $6 + 5 = 11$ people. Since the order does not matter, we use combinations. Step 1: Calculate the total number of people. Total people = $6 men + 5 women = 11 \text{ people}$

Step 2: Use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ to choose 4 people from 11. $\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!}$ $= \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!}$ $= \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}$ $= 11 \times 10 \times 3 = 330$ The number of ways is 330.

b) the committee has 2 men and 2 women We need to choose 2 men from 6 men AND 2 women from 5 women. Step 1: Choose 2 men from 6 men. $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15$ Step 2: Choose 2 women from 5 women. $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10$ Step 3: Multiply the number of ways to choose men by the number of ways to choose women. $15 \times 10 = 150$ The number of ways is 150.

c) the committee has only 1 woman If the committee has 1 woman, and it's a 4-person committee, then the remaining $4 - 1 = 3$ members must be men. Step 1: Choose 1 woman from 5 women. $\binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = \frac{5 \times 4!}{1 \times 4!} = 5$ Step 2: Choose 3 men from 6 men. $\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = \frac{120}{6} = 20$ Step 3: Multiply the number of ways to choose women by the number of ways to choose men. $5 \times 20 = 100$ The number of ways is 100.