I'm having trouble on how to figure something out. The question says a pizza place uses 5 toppings (a,b,c,d,e). How many different combinations can be made with the pizzas having 3 different topping? A

It's a question about factorials: the answer is the number of ways you can choose 3 from 5. You could enumerate them without too much difficulty, but the shortcut is to work out 5! / (3! x (5-3)!), where N! means N x (N-1) x (N-2) .... x 2 x 1.

think of it this way:

The pizzamaker would have 5 ways to choose the first topping, then 4 ways to do the second topping and finally 3 ways for the third topping

so isn't that 5x4x3 or 60 ways

No - you're distinguishing between the 1st, 2nd and 3rd, which you don't need to do. ABC is the same as BCA, which is the same as CAB etc. You've got to allow for all the combinations that are the same.

In fact, every triple can occur in six different ways - so the answer is actually the 60 you identified, divided by 6.

David is correct, I was in error

My senior moment of the day.

I'm in 6th grade and never heard of using ! in math. I guess I'll do it the long way to see how it works. Thanks so much :)

do 5x3..

its fifteen(:
these ppl are so confusingg;

Just for the record the answer is ten. They are: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE and CDE.

To find the number of different combinations that can be made with the pizzas having 3 different toppings out of 5 possible toppings (a,b,c,d,e), you can use the formula for combinations.

The formula for combinations is represented as "n choose r" and can be calculated using the formula:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of items to choose from and r is the number of items you want to choose.

In this case, n = 5, as there are 5 possible toppings (a,b,c,d,e), and r = 3, as you want to choose 3 different toppings for the pizzas.

Now, substitute these values into the formula:

C(5, 3) = 5! / (3! * (5-3)!)

Calculating the factorial terms:

5! = 5 * 4 * 3 * 2 * 1 = 120
3! = 3 * 2 * 1 = 6
(5-3)! = 2! = 2 * 1 = 2

Substituting these values back into the formula:

C(5, 3) = 120 / (6 * 2) = 10

Therefore, there are 10 different combinations that can be made with pizzas having 3 different toppings out of the 5 possible toppings (a,b,c,d,e).