1. ## Combinations

Ballots for municipal elections usually list candidates for several different positions. If a resident can vote for a mayor, two councillors, a school trustee, and a hydro commissioner, how many combinations of positions could the resident choose to mark on one ballot?

(2x3x2x2)-1
=24-1
=23?

The answer in in the book says 31 but i don't understand how or if the book is wrong.

2. Originally Posted by KobeBryant
Ballots for municipal elections usually list candidates for several different positions. If a resident can vote for a mayor, two councillors, a school trustee, and a hydro commissioner, how many combinations of positions could the resident choose to mark on one ballot?

(2x3x2x2)-1
=24-1
=23?

The answer in in the book says 31 but i don't understand how or if the book is wrong.
I think you are not saying the whole story. For example, two councillors out of how many?

3. That's the full question copied word for word, I'm thinking that the back of the book is wrong.

4. It seems the question is asking for the different ways the voter could vote or not vote for the 5 different positions. For example, the voter could vote for only mayor and hydro commissioner. That's 2 positions. There are $\binom{5}{2} = 10$ ways to vote for 2 positions. The total number of ways to vote is

$\sum_{k=1}^5 \binom{5}{k} = 5 + 10 + 10 + 5 + 1 = 31.$

5. Just wanted to remind you a useful formula:

$
\sum_{k=0}^n \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}1^k1^{n-k}= (1+1)^n=2^n.
$

So $
\sum_{k=1}^5 \binom{5}{k} = 2^5-1 = 31.
$

6. Originally Posted by Rebesques
Just wanted to remind you a useful formula:

$
\sum_{k=0}^n \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}1^k1^{n-k}= (1+1)^n=2^n.
$
I did not know this. Thanks for pointing it out.

7. Originally Posted by Rebesques
Just wanted to remind you a useful formula:

$
\sum_{k=0}^n \binom{n}{k} = \sum_{k=0}^n \binom{n}{k}1^k1^{n-k}= (1+1)^n=2^n.
$
I like that one!

-Dan