Sure, let's break down each of the questions and solve them step-by-step:
### Question 1:
An ice cream shop has a choice of 10 toppings. In how many ways can you choose 3 different toppings for your ice cream?
To solve this, we need to use the concept of combinations, where the order of selection does not matter. The number of ways to choose [tex]\(r\)[/tex] items out of [tex]\(n\)[/tex] items is given by the combination formula:
[tex]\[ C(n, r) = \frac{n!}{r!(n - r)!} \][/tex]
Here, [tex]\(n = 10\)[/tex] and [tex]\(r = 3\)[/tex].
[tex]\[ C(10, 3) = \frac{10!}{3!(10 - 3)!} = \frac{10!}{3!7!} = 120.0 \][/tex]
So, there are 120 ways to choose 3 different toppings out of 10.
### Question 2:
Out of a group of 30 people, 20 of which are women, how many different committees of 6 people can be chosen?
Again, we use the combination formula since we are choosing 6 people out of 30 without regard to order. Here, [tex]\(n = 30\)[/tex] and [tex]\(r = 6\)[/tex].
[tex]\[ C(30, 6) = \frac{30!}{6!(30 - 6)!} = \frac{30!}{6!24!} = 593775.0 \][/tex]
So, there are 593,775 different ways to form a committee of 6 people from a group of 30.
### Question 3:
In Exercise 2, determine the possible number of committees of 6 in which 4 are women and 2 are men.
Here, we need to split the problem into two parts: selecting the women and selecting the men.
- First, we choose 4 women out of 20.
- Next, we choose 2 men out of the remaining (30 - 20) = 10 men.
For selecting 4 women out of 20:
[tex]\[ C(20, 4) = \frac{20!}{4!(20 - 4)!} = \frac{20!}{4!16!} = 4845 \][/tex]
For selecting 2 men out of 10:
[tex]\[ C(10, 2) = \frac{10!}{2!(10 - 2)!} = \frac{10!}{2!8!} = 45 \][/tex]
Now, multiply these two results to get the total number of ways to form such a committee:
[tex]\[ 4845 \times 45 = 218025.0 \][/tex]
So, there are 218,025 possible committees of 6 people where 4 are women and 2 are men.
