1. Probability

If you have 4 flavors (chocolate, cookies-n-cream, strawberry and vanilla) of ice cream and 6 toppings (caramel, hot fudge, marshmallow, M&Ms, nuts and strawberries), how many sundaes are possible using one flavor of ice cream and three different toppings?

My work: (6)(5)(4)=120

How many sundaes are possible using one flavor of ice cream and from 0 to 6 toppings?

My work: (6)(5)(4)(3)(2)(1)+(6)(5)(4)(3)(2)+(6)(5)(4)(3)+(6 )(5)(4)+(6)(5)+6+1(<---for 0 toppings?)=1957 I am really unsure about how I did this part

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

My work: (4)(4)(4)=64

Does anyone know if I am on the right track?

2. Originally Posted by antman
If you have 4 flavors (chocolate, cookies-n-cream, strawberry and vanilla) of ice cream and 6 toppings (caramel, hot fudge, marshmallow, M&Ms, nuts and strawberries), how many sundaes are possible using one flavor of ice cream and three different toppings?

My work: (6)(5)(4)=120

How many sundaes are possible using one flavor of ice cream and from 0 to 6 toppings?

My work: (6)(5)(4)(3)(2)(1)+(6)(5)(4)(3)(2)+(6)(5)(4)(3)+(6 )(5)(4)+(6)(5)+6+1(<---for 0 toppings?)=1957 I am really unsure about how I did this part

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

My work: (4)(4)(4)=64

Does anyone know if I am on the right track?
(a) Each sundae must comprise one and only one flavor of ice cream. There are 4 different flavors; therefore, every topping combo will be multiplied by 4. The number of topping combos are given by 6 choose 3.

(b) 4 *((6 choose 0) + (6 choose 1) + (6 choose 2) + ......(6 choose 6))

(c) Correct.

3. Originally Posted by antman
[snip]
How many different combinations of flavors of three scoops of ice cream are possible if it is permissible to make all three scoops the same flavor?

My work: (4)(4)(4)=64

Does anyone know if I am on the right track?
You didn't say if the order of the scoops is significant or not. If the order matters, your answer is correct. if not, there are
$\binom{4+3-1}{3} = 20$
possible arrangements.

4. So my answer for a would just be multiplied by 4 [4*(6)(5)(4)=4*120=480] to get 480?

For part b would also just be multiplied by 4 [4*(1+(6)(5)(4)(3)(2)+(6)(5)(4)(3)+(6)(5)(4)+(6)(5) +(6)+(1))=4*1238=4952 to get 4952? I replaced the first term with 1 because there is only one way to choose six toppings. I'm hoping this question is asking for 0 to 6 DIFFERENT toppings, but I am not sure.

a) 480
b) 4952

5. I think I figured this problem out.
My answer for a would be 4*(6!/3!3!)=4*20=80?
For part b, my answer would be 4(1+6+15+20+15+6+1)=4*64=256?

6. Originally Posted by awkward
You didn't say if the order of the scoops is significant or not. If the order matters, your answer is correct. if not, there are
$\binom{4+3-1}{3} = 20$
possible arrangements.
This problem (where the order of scoops does not matter) is an example of "combinations with repetition". In general, if you have n things to choose from and you are going to pick r items, the number of ways this can be done is

$\binom{n+r-1}{r} = \frac{(n+r-1)!}{r! \; (n-1)!}$.

You can find a derivation of this very useful formula here: CSCE 235 Discrete Math Project - Combinations with Repetition