Number of possible ways

[BigJellyFish]

Hello,
I have such an exercise which I find hard to solve. It is : "Everyday Paul buys either an ice cream for 1 pound or a chocolate for 2 pounds. He can choose from 4 tastes of ice creams, but there is only one type of chocolate he can choose. In how many ways can Paul spend x pounds?" I was trying to find a pattern between the first x=1 - 4 ways, x=2 - 17 ways etc but it is hard - I guess the 4ⁿ will be in the formula, but then I can't think of the rest.
Thanks in advance for help

 

[pka]

It is : "Everyday Paul buys either an ice cream for 1 pound or a chocolate for 2 pounds. He can choose from 4 tastes of ice creams, but there is only one type of chocolate he can choose. In how many ways can Paul spend x pounds?" I was trying to find a pattern between the first x=1 - 4 ways, x=2 - 17 ways etc but it is hard - I guess the 4ⁿ will be in the formula, but then I can't think of the rest.

If \(\displaystyle \bf X\) is the number of pounds one can spend on ice cream then there are \(\displaystyle \dfrac{(X+3)!}{X!(3!)}\) ways to do that.
To explain: If we are to choose \(\displaystyle N\) items from \(\displaystyle K\) kinds it can be done in \(\displaystyle \dfrac{(N+K-1)!}{N!(K-1)!}\) to do that.
In the above \(\displaystyle K=4\) because there four flavors of creams.
Suppose Paul is to choose \(\displaystyle \bf C\) units of chocolates that would be \(\displaystyle 2C\) pounds.

If Paul has \(\displaystyle X\) pounds to spend, he can do that if \(\displaystyle X=N+2C\) and that can be done \(\displaystyle \dfrac{(N+3)!}{N!(3!)}\) ways.
Note that if \(\displaystyle N=0\) then \(\displaystyle C\) must be even else Paul has a pound left over.

 

[BigJellyFish]

Thanks for your answer pka !