The student council is ordering pizza for their next meeting. There are 20 council members, 7 of whom are vegetarian. A committee of 3 will order 6 pizzas from a pizza shop that has a special price for large pizzas with up to three toppings. The shop offers 10 different toppings.
a)How many different pizza committees can the council choose if there must be at least 1 vegitarian and 1 non-vegitarian on the committee?
b)In how many ways could the committee choose up to 3 toppings for a pizza?
c)The committee wants as much variety as possible in the toppings. They decide to order each topping exactally once to have at least 1 topping on each pizza. Describe the different cases possible when distributing the toppings in this way.
d)For one of these cases, determine the number of ways of choosing and distributing the 10 toppings.
a) Well, well, well, pizza committees! Sounds like a delicious problem. Now, since you must have at least 1 vegetarian and 1 non-vegetarian in the committee, let's break it down. You have 7 vegetarians and 13 non-vegetarians to choose from, and you need to form a committee of 3. We can calculate the number of committees using combinations: C(7, 1) * C(13, 2). That gives us 7 * 78 = 546 different pizza committees. That's a lot of mouths to feed!
b) Now let's talk toppings! There are 10 different toppings to choose from, and the committee can choose up to 3 toppings for each pizza. So, we have to figure out the number of ways we can choose 0, 1, 2, or 3 toppings. Let's put on our math hats and calculate those possibilities. We'll use the binomial coefficient formula, also known as combinations. The total number of ways to choose up to 3 toppings is C(10, 0) + C(10, 1) + C(10, 2) + C(10, 3). Grab your calculator and let's compute that sum!
c) Ah, the committee wants variety in their toppings. How fancy! Let's think about the different cases here. We know we need at least 1 topping on each pizza, so we can distribute the toppings in the following ways:
- Case 1: Each pizza has 1 topping. This means we need 6 different toppings. How many ways can we choose and distribute those 6 toppings?
- Case 2: Two pizzas have 1 topping each, and the remaining four have 2 toppings each. How many ways can we choose the 2 toppings for each pizza and distribute them?
- Case 3: Three pizzas have 1 topping each, and the remaining three have 3 toppings each. How many ways can we choose the 3 toppings for each pizza and distribute them?
d) Let's focus on one case to make things a bit more manageable. How about Case 1, where each pizza has just 1 topping? We have 10 different toppings to choose from, and we need to distribute them among 6 pizzas. To calculate the number of ways, we can use the concept of permutations (remember, order matters!). The formula to calculate the number of permutations is nPr = n! / (n - r)!, where n is the total number of items and r is the number of items to be chosen. Now go ahead and enjoy the toppings-permuting process!
a) To find the number of different pizza committees the council can choose, we need to consider the requirements: at least 1 vegetarian and 1 non-vegetarian member on the committee.
Step 1: Select the vegetarian member. Given that there are 7 vegetarian council members, we can choose 1 from those 7, which can be done in C(7, 1) ways.
Step 2: Select the non-vegetarian member. Now, we have 19 council members remaining (20 - 1 vegetarian member). Out of these, we can select 1 non-vegetarian member in C(19, 1) ways.
Step 3: Select the remaining member. After selecting 1 vegetarian and 1 non-vegetarian member, we need to select 1 more member for the committee. From the remaining members (18 after the previous selections), we can choose 1 member in C(18, 1) ways.
Step 4: Multiply the number of ways together from steps 1, 2, and 3 to get the total number of different pizza committees:
C(7, 1) * C(19, 1) * C(18, 1) = 7 * 19 * 18 = 2394
Therefore, there are 2394 different pizza committees the council can choose.
b) To determine the number of ways the committee can choose up to 3 toppings for a pizza, we need to consider the shop's special offer of large pizzas with up to three toppings.
Step 1: Select the number of toppings for the first pizza. We have three options: 0 toppings, 1 topping, or 2 toppings. Therefore, we have 3 choices.
Step 2: Select the number of toppings for the second pizza. Similar to the first pizza, we have 3 choices.
Step 3: Select the number of toppings for the third pizza. Again, we have 3 choices.
Step 4: Multiply the number of choices from each step together to get the total number of ways to choose up to 3 toppings for a pizza:
3 * 3 * 3 = 3^3 = 27
Therefore, there are 27 ways the committee can choose up to 3 toppings for a pizza.
c) In this case, the committee wants to have each topping used exactly once, with at least 1 topping on each pizza. Let's consider the different scenarios.
Scenario 1: Each pizza has exactly 1 topping.
The committee can arrange the toppings in the following ways:
- Pizza 1: One of the 10 toppings
- Pizza 2: One of the 9 remaining toppings
- Pizza 3: One of the 8 remaining toppings
So, there are 10 * 9 * 8 = 720 ways to distribute the toppings in this scenario.
Scenario 2: Two pizzas have exactly 1 topping, while one pizza has 2 toppings.
The committee can arrange the toppings in the following ways:
- Pizza 1: One of the 10 toppings
- Pizza 2: One of the 9 remaining toppings
- Pizza 3: Two of the 8 remaining toppings (C(8, 2) ways)
So, there are 10 * 9 * C(8, 2) ways to distribute the toppings in this scenario.
Scenario 3: One pizza has exactly 1 topping, while two pizzas have 2 toppings each.
The committee can arrange the toppings in the following ways:
- Pizza 1: One of the 10 toppings
- Pizza 2: Two of the 9 remaining toppings (C(9, 2) ways)
- Pizza 3: Two of the 7 remaining toppings (C(7, 2) ways)
So, there are 10 * C(9, 2) * C(7, 2) ways to distribute the toppings in this scenario.
Scenario 4: One pizza has exactly 2 toppings, while two pizzas have 3 toppings each.
The committee can arrange the toppings in the following ways:
- Pizza 1: Two of the 10 toppings (C(10, 2) ways)
- Pizza 2: Three of the 8 remaining toppings (C(8, 3) ways)
- Pizza 3: Three of the 5 remaining toppings (C(5, 3) ways)
So, there are C(10, 2) * C(8, 3) * C(5, 3) ways to distribute the toppings in this scenario.
d) To determine the number of ways of choosing and distributing the 10 toppings in one of these cases, let's choose Scenario 1. In this case, each pizza has exactly 1 topping.
The committee can distribute the toppings in the following ways:
- Pizza 1: One of the 10 toppings
- Pizza 2: One of the 9 remaining toppings
- Pizza 3: One of the 8 remaining toppings
So, there are 10 * 9 * 8 = 720 ways to choose and distribute the 10 toppings in this case.