help with combinations

[kory]

Alis buying soda for his son’s birthday party. He needs to purchase 10 2-liter bottles of soda. He stops at a convenience store on the way home and finds that they have 17 different types of soda. Assuming the store stocks enough of each flavor that there are enough for him to get at least 10 of any type, how many different selections of 10 bottles could Al make if...


a.)There are no restrictions on the flavors of soda he purchases?
Solution: [math] 17^{10}= 2015993900449 [/math]

b.)He wants to purchase at least 2 root beer?
Solution:
case for no rootbeer:[math] 16^{10}=1099511627776[/math]case for 1 rootbeer:[math] 16^{9}=68719476736[/math]
I believe I add these two together and subtract that from[math] 17^{10}[/math] however I could be wrong...


c.)He wants to purchase at most 3 cola?

Im not sure how to handle c.)...any advice?

 

[pka]

Alis buying soda for his son’s birthday party. He needs to purchase 10 2-liter bottles of soda. He stops at a convenience store on the way home and finds that they have 17 different types of soda. Assuming the store stocks enough of each flavor that there are enough for him to get at least 10 of any type, how many different selections of 10 bottles could Al make if...
a.)There are no restrictions on the flavors of soda he purchases?
b.)He wants to purchase at least 2 root beer?
c.)He wants to purchase at most 3 cola?

Your answers are not correct. This question is about putting \(N\) identical objects (in this case choices) into \(K\) distinct cells (in this case flavors).
That can be done is \(\dbinom{N+K-1}{N}=\dfrac{(N+K-1)!}{N!(K-1)!}\).
So for part a) \(\dbinom{10+17-1}{10}\).

 

[kory]

Hmmmm.....I thought it was asking for all of the potential flavors that could be purchased....well ok.
For b.) i think I need to consider zero root bears and one rootbeer. So would my formula be [math] \binom {10+16-1}{10}[/math] + [math] \binom {9+16-1}{9}[/math] and then subtract that from a.)