Unlike a permutation, in a combination the order of the objects selected does not matter. For example, if you were choosing a team of three students from a class of ten, the order you said "Joe, Sally, John" would not matter to who is in the group- saying "Sally, Joe, John" instead doesn't change anything.

The number of ways you can make a combination of r objects out of a set of n objects is made from the formula below:

C=n!/r!(n-r)!

You may be asking (like I did when I learned this material), "What? That formula doesn't make sense!".- However, it is based on the formula for a permutation. If you think about it, the n! makes perfect sense- thats how many options you'd have if you picked in order from all of them. Removing the (n-r)! gets it down to just the numbers you want (n-r equals the last number you arn't picking. If you pick 3 from 5, it'll be 5!/(5-3)!, so 5*4*3*2*1/2*1, which, after cancelling it out to 5*4*3, accomplishes the "pick x of 5" part.) The extra r! removes all the redundant options. For example, say you have 6 different balls, labeled A through F. If you pick three, you can pick ABC, ABD, ABE, ABF, etc. You'll end up with six of the same- ABC, ACB, BAC, BCA, CAB, CBA- all of which only count as one in a combination. As you'll notice, since we picked 3, 3!=3*2*1=6- the same number we have to divide out!

Lets take the example above. Out of a class of ten students, how many ways could you make a team of three students? We know n is ten because the set (the students) has ten objects in it. We know r is 3 because the team will have three students on it. To solve, first we set up the equation:

C=n!/r!(n-r)!

Then we fill in the values:

C=10!/3!(10-3)!

C=10!/3!7!

By taking apart the factorials, we can simplify the 10! and 7!. Alternativley, just start at 10 and multiply down, stopping right before 7.

C=10*9*8*7*6*5*4*3*2*1/3*2*1*7*6*5*4*3*2*1

C=10*9*8/3*2*1

C=720/6

C= 120

There you have it, there are 120 ways to pick a team of 3 people out of ten.

## How to solve a combination

The number of ways you can make a combination of r objects out of a set of n objects is made from the formula below:

C=n!/r!(n-r)!

## An Explanation

You may be asking (like I did when I learned this material), "What? That formula doesn't make sense!".- However, it is based on the formula for a permutation. If you think about it, the n! makes perfect sense- thats how many options you'd have if you picked in order from all of them. Removing the (n-r)! gets it down to just the numbers you want (n-r equals the last number you arn't picking. If you pick 3 from 5, it'll be 5!/(5-3)!, so 5*4*3*2*1/2*1, which, after cancelling it out to 5*4*3, accomplishes the "pick x of 5" part.) The extra r! removes all the redundant options. For example, say you have 6 different balls, labeled A through F. If you pick three, you can pick ABC, ABD, ABE, ABF, etc. You'll end up with six of the same- ABC, ACB, BAC, BCA, CAB, CBA- all of which only count as one in a combination. As you'll notice, since we picked 3, 3!=3*2*1=6- the same number we have to divide out!

## An Example of a Combination

Lets take the example above. Out of a class of ten students, how many ways could you make a team of three students? We know n is ten because the set (the students) has ten objects in it. We know r is 3 because the team will have three students on it. To solve, first we set up the equation:

C=n!/r!(n-r)!

Then we fill in the values:

C=10!/3!(10-3)!

C=10!/3!7!

By taking apart the factorials, we can simplify the 10! and 7!. Alternativley, just start at 10 and multiply down, stopping right before 7.

C=10*9*8*7*6*5*4*3*2*1/3*2*1*7*6*5*4*3*2*1

C=10*9*8/3*2*1

C=720/6

C= 120

There you have it, there are 120 ways to pick a team of 3 people out of ten.

Combinations
Reviewed by Ramandeep Singh
on
Saturday, March 01, 2014
Rating:

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