Here are the solutions for Exercise 1: Exercise 1: How many ways can you form a 4-person committee from 6 men and 5 women? Total number of people = 6 men + 5 women = 11 people. The committee size is 4 people. a) With no restrictions Step 1: Determine the total number of people ($$n$$) and the committee size ($$k$$). $$n = 11$$ (total people) $$k = 4$$ (committee members) Step 2: Use the combination formula, as the order of selection does not matter for a committee. $$C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$$ Step 3: Substitute the values and calculate. $$ C(11, 4) = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} $$ $$ C(11, 4) = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} $$ $$ C(11, 4) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} $$ $$ C(11, 4) = \frac{7920}{24} $$ $$ C(11, 4) = 330 $$ The number of ways to form a 4-person committee with no restrictions is $$\boxed{330}$$. b) the committee has 2 men and 2 women Step 1: Determine the number of ways to choose 2 men from 6 men. $$ C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} $$ $$ C(6, 2) = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15 $$ Step 2: Determine the number of ways to choose 2 women from 5 women. $$ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} $$ $$ C(5, 2) = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10 $$ Step 3: Multiply the number of ways to choose men and women, as these are independent selections. Total ways = $$C(6, 2) \times C(5, 2) = 15 \times 10 = 150$$ The number of ways to form a committee with 2 men and 2 women is $$\boxed{150}$$. c) the committee has only 1 woman Step 1: If the committee has only 1 woman, and the committee size is 4, then the remaining 3 members must be men. Determine the number of ways to choose 1 woman from 5 women. $$ C(5, 1) = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} $$ $$ C(5, 1) = \frac{5 \times 4!}{1 \times 4!} = 5 $$ Step 2: Determine the number of ways to choose 3 men from 6 men. $$ C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} $$ $$ C(6, 3) = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = \frac{120}{6} = 20 $$ Step 3: Multiply the number of ways to choose women and men. Total ways = $$C(5, 1) \times C(6, 3) = 5 \times 20 = 100$$ The number of ways to form a committee with only 1 woman is $$\boxed{100}$$.