1. ## combinatorial voting problem

1) how many ways can you vote if there are 4 candidates you get 12 votes but you are not required to use all 12.

x1+x2+x3+x4+x5 = 12
C(12+5,12)= 6,188
is this correct?

2) how many different dinners can be created if
4 choices of pasta(only can choose 1), 5 choices of sauces(only can choose 1),2 choices of salads(only can choose 1), 4 choices of meats(can choose as many as you want assume they are different ie meat meat isnt allowed), 5 additional toppings(choose up to 3)

2. ## Combination with repetition

Hello canyiah
Originally Posted by canyiah
1) how many ways can you vote if there are 4 candidates you get 12 votes but you are not required to use all 12.

x1+x2+x3+x4+x5 = 12
C(12+5,12)= 6,188
is this correct?
I think this should be $\binom{16}{12} = 1820$.

The problem is equivalent to choosing $12$ items from $5$, with repetition being allowed. The $12$ items are the $12$ votes, each for a particular candidate or none; the $5$ are the $4$ candidates plus a 'null' vote, where the vote is not cast for any of the candidates.

Now the number of ways of choosing $k$ items from $n$, with repetition allowed, is:
$\binom{n+k-1}{k}$ (See here, for example.)
So when $k = 12$ and $n = 5$, this is:
$\binom{16}{12} = 1820$
2) how many different dinners can be created if
4 choices of pasta(only can choose 1), 5 choices of sauces(only can choose 1),2 choices of salads(only can choose 1), 4 choices of meats(can choose as many as you want assume they are different ie meat meat isnt allowed), 5 additional toppings(choose up to 3)
Since repetition is not allowed in the choice of meat, I'm assuming that it is not allowed in the choice of additional topping either. I'm also assuming that you must choose exactly 1 pasta, 1 sauce and 1 salad, but that you can choose 0 meats and 0 additional toppings (since the phrases 'as many as you want' and 'up to' are used for these items).

Then we have:
Number of choices of pasta: $4$

Number of choices of sauce: $5$

Number of choices of salad: $2$

Number of choices of meat: $2^4=16$, since each of the $4$ meats may be either chosen or rejected

Number of choices of additional topping: $\binom50+\binom51+\binom52+\binom53=26$
So, if my assumptions are correct, I reckon the number of different dinners is:
$4\times5\times2\times16\times26=16640$