sal_xcool wrote:If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, how many different combinations of candidates could be elected to the Board?

A. 6
B. 8
C. 12
D. 16
E. 24

OA D

When I looked at this question, I thought I have to use the combination formula since it's asking for the number of combinations and as the result I got A.

4 positions and 2 candidates -
1st position can be filled in 2 ways
2nd position can be filled in 2 ways
3rd position can be filled in 2 ways
4th position can be filled in 2 ways

So, the number of ways in which the 4 positions can be filled in by the 2 candidates = 2*2*2*2 = 16 D

sal_xcool wrote:

Can someone explain to me once for all when to use the combination equation and when to use the permutation equation?

Thank you in advance.

combination -> when order doesn't matter
permutation -> when order matters

Let's say you're going to pick two people from a group of 3 people : A, B, C.
If nothing else is mentioned then it doesn't matter if you pick A first and B second or vice versa. This is an example of combinations. There are 3 possible combinations - AB, AC and BC.

If you now say, the first person is going to be given a gold medal and the second a silver one; now order counts.
So picking AB is different than BA, and now there are six possibilities (permutations) AB, AC, BA, BC etc.

But please note that you can't directly apply the permutation formula in the above problem you mentioned (i.e. the answer wouldn't be 4P2), as here we are not actually arranging the 2 candidates to the 4 chairs. In the given problem , it is possible that a single candidate can be chosen for all the 4 chairs (which is not taken care by the permutation formula).

Hope it helps a bit.

anshumishra wrote:

sal_xcool wrote:

Can someone explain to me once for all when to use the combination equation and when to use the permutation equation?

Thank you in advance.

combination -> when order doesn't matter
permutation -> when order matters

Let's say you're going to pick two people from a group of 3 people : A, B, C.
If nothing else is mentioned then it doesn't matter if you pick A first and B second or vice versa. This is an example of combinations. There are 3 possible combinations - AB, AC and BC.

If you now say, the first person is going to be given a gold medal and the second a silver one; now order counts.
So picking AB is different than BA, and now there are six possibilities (permutations) AB, AC, BA, BC etc.

But please note that you can't directly apply the permutation formula in the above problem you mentioned (i.e. the answer wouldn't be 4P2), as here we are not actually arranging the 2 candidates to the 4 chairs. In the given problem , it is possible that a single candidate can be chosen for all the 4 chairs (which is not taken care by the permutation formula).

Hope it helps a bit.

Thanks, I guess the difficult thing for me is to see when to use the combination or permutation or just multiply all the possibilities as it's the case in this question. good explanation.

sal_xcool wrote:If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, how many different combinations of candidates could be elected to the Board?

A. 6
B. 8
C. 12
D. 16
E. 24

OA D

When I looked at this question, I thought I have to use the combination formula since it's asking for the number of combinations and as the result I got A.

Can someone explain to me once for all when to use the combination equation and when to use the permutation equation?

Thank you in advance.

Hi!

Good explanations above, I just wanted to raise one important math concept - the fundamental counting principle.

Here's the rule:

when there are m ways for event A to occur, and n ways for event B to occur, if events A and B are independent, then there are m*n ways for both A and B to occur.

In other words:

when calculating the number of ways that multiple independent events can occur, you MULTIPLY the individual possibilities.

Here, we have 4 independent events, each of which can occur in 2 ways, so the answer is simply:

2*2*2*2 = 16