Statistics question (combinations/permutations)

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Homework Statement



The eating club is hosting a make your own sundae at which the following are provided:

Ice cream flavors: chocolate, cookies-n-cream, strawberry, vanilla

Toppings: caramel, hot fudge, marshmellow, m & m's, nuts, strawberries

a) How many sundaes are possible using one flavor of ice cream and from zero to six toppings?

b) How many different combinations of flavors of three scoops of ice cream are possible if it is permissable to make all three scoops the same flavor?

Homework Equations



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

n P r = n! / (n-r)!

The Attempt at a Solution



a) I know the answer is 256. However, I'm not sure how to get to it.

I know the first part of the answer requires 4 C 1, however I'm not sure how to arrive at 4 * 64.

b) Answer is 20. They are asking for something equivalent to 4 C 3, which would be 4. However, I am guessing since it is possible to use the same flavor, the answer is multiplied 4 times, plus the 4 sets of the same flavors (i.e. [(4 C 3) * 4] + 4).
 
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  • #2
For part a, there are two steps here. First, you have to choose a flavor of icecream. As you said, there are 4 C 1 possibilities, since we can only choose 1 flavor. Now look at the toppings. We want to find the total number of combinations of toppings. There's a couple of different ways of getting 64, but this is the way I look at it. Look at this as a 6 digit binary word, where the slot correspondings to the type of topping, 1 means that you have that topping and 0 means that you don't. For instance:

011001: no caramel, hot fudge, marshmallow, no m&m's, no nuts, and strawberries.

Knowing this now, what's the total number of 6 digit binary words? So, what's the total number of possible toppings?
 
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  • #3
gb7nash said:
For part a, there are two steps here. First, you have to choose a flavor of icecream. As you said, there are 4 C 1 possibilities, since we can only choose 1 flavor. Now look at the toppings. We want to find the total number of combinations of toppings. There's a couple of different ways of getting 64, but this is the way I look at it. Look at this as a 6 digit binary word, where the slot correspondings to the type of topping, 1 means that you have that topping and 0 means that you don't. For instance:

011001: no caramel, hot fudge, marshmallow, no m&m's, no nuts, and vanilla.

Knowing this now, what's the total number of 6 digit binary words? So, what's the total number of possible toppings?

Cool got it. So it's 26 * 4.

Still not sure about b).
 
  • #4
If only one flavor is allowed, all three scoops have the same flavor. What's the number of ways you can do this?
 
  • #5
gb7nash said:
If only one flavor is allowed, all three scoops have the same flavor. What's the number of ways you can do this?

Ok, so if you have the same flavor you can do it four different ways.

But I'm not understanding how you take the other combinations into account.
 
  • #6
Do we say (4 C 3) for each flavor, so we multiply the result (4) by 4, and then add the same flavors (4)...

So the answer is (4*4) + 4?
 
  • #7
trojansc82 said:
Do we say (4 C 3) for each flavor, so we multiply the result (4) by 4, and then add the same flavors (4)...

So the answer is (4*4) + 4?

I'm not sure why you're doing this. If we're only talking about flavors of icecream, the answer is just 4. There are 4 different combinations that give you the same flavors of icecream, e.g. Vanilla Vanilla Vanilla, Chocolate Chocolate Chocolate, ...
 
  • #8
gb7nash said:
I'm not sure why you're doing this. If we're only talking about flavors of icecream, the answer is just 4. There are 4 different combinations that give you the same flavors of icecream, e.g. Vanilla Vanilla Vanilla, Chocolate Chocolate Chocolate, ...

Yeah, it's permissable to have the same flavor for all three, but you have to take into account the combination of the others.

So having the same flavor for three scoops is 4, but it has to be added to the others.
 
  • #9
The number of combinations of r items selected from n, allowing for repetition, is (n-1)+rCr

with that, 6C3=20
 

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