Combinations

**KobeBryant** · May 1st 2006, 06:28 PM

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.

**ThePerfectHacker** · May 1st 2006, 06:53 PM

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?

**KobeBryant** · May 1st 2006, 07:40 PM

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

**JakeD** · May 1st 2006, 08:00 PM

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

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

**Rebesques** · May 2nd 2006, 11:13 AM

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. $

**JakeD** · May 6th 2006, 02:13 PM

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.

**topsquark** · May 6th 2006, 05:10 PM

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

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