College Algebra
Tutorial 57: Combinations
Learning Objectives
Introduction
In this tutorial we will be going over combinations. When you need to count the number of groupings, without regard to order, then combinations are the way to go. Recall that permutations specifically count the number of ways a task can be arranged or ordered. That is the difference between the two, permutations is with regard to order and combinations is without regard to order. If you need a review on permutations, feel free to go to Tutorial 56: Permutations. Let's see what you can do with these combinations.
Tutorial
Combination
An arrangement of r objects,
WITHOUT regard to ORDER and without repetition,
selected from n distinct objects is called
a
combination of n objects taken r at
a time.
The number of such combinations is denoted by
The n and the r mean the same thing in both the permutation and combinations, but the formula differs. Note that the combination has an extra r! in its denominator.
If you need review on permutations or factorials, feel free to go to Tutorial
56: Permutations.
Note that if we were putting these teams in any kind of order, then we would need to use permutations to solve the problem.
But in this case, order does not matter, so we are going to use combinations.
First we need to find n and r :
If you said n = 9 you are correct!!! There are 9 teams in the conference.
If r is the number of teams we are using at a time, what do you think r is?
If you said r = 2, pat yourself on
the back!! 2 teams play per game.
Let's put those values into the combination formula and see what
we get:
*Eval. inside ( )
*Expand 9! until it gets to 7!
which is the larger ! in the den.
*Cancel out 7!'s
If you don't have a factorial key, you can simplify it as shown above
and then enter it in. It is probably best to simplify it first, because
in some cases the numbers can get rather large, and it would be cumbersome
to multiply all those numbers one by one.
Wow, this means there are 36 different games in the conference.
Note that if we were putting these cards in any kind of order, then we would need to use permutations to solve the problem.
But in this case, order does not matter, so we are going to use combinations.
First we need to find n and r :
If you said n = 52 you are correct!!! There are 52 cards in a deck of cards.
If r is the number of cards we are using at a time, what do you think r is?
If you said r = 4, pat yourself on
the back!! We want 4 card hands.
Let's put those values into the combination formula and see what
we get:
*Eval. inside ( )
*Expand 52! until it gets to 48!
which is the larger ! in the den.
*Cancel out 48!'s
If you don't have a factorial key, you can simplify it as shown above
and then enter it in. It is probably best to simplify it first, because
in some cases the numbers can get rather large, and it would be cumbersome
to multiply all those numbers one by one.
Wow, this means there are 270,725 different 4 card hands.
Note that there are no special conditions placed on the marbles that we draw, so this is a straight forward combination problem.
Note that if we were putting these marbles in any kind of order, then we would need to use permutations to solve the problem.
But in this case, order does not matter, so we are going to use combinations.
First we need to find n and r:
If you said n = 8 you are correct!!! There are 3 red and 5 white marbles for a total of 8 marbles.
If r is the number of marbles we are drawing at a time, what do you think r is?
If you said r = 3, pat yourself on
the back!! 3 marbles are drawn at a time.
Let's put those values into the combination formula and see what
we get:
*Eval. inside ( )
*Expand 8! until it gets to 5!
which is the larger ! in the den.
*Cancel out 5!'s
If you don't have a factorial key, you can simplify it as shown above
and then enter it in. It is probably best to simplify it first, because
in some cases the numbers can get rather large, and it would be cumbersome
to multiply all those numbers one by one.
Wow, this means there are 56 different draws.
In part a above, we looked at all possible draws. From that list we only want the ones that contain only red.
Let's see what the draw looks like: we would have to have 3 red marbles to meet this condition:
3 RED
First we need to find n and r :
If you said n = 3 you are correct!!! There are a total of 3 red marbles.
If r is the number of RED marbles we are drawing at a time, what do you think r is?
If you said r = 3, pat yourself on
the back!! 3 RED marbles are drawn at a time.
Let's put those values into the combination formula and see what
we get:
*Eval. inside ( )
*Cancel out 3!'s
If you don't have a factorial key, you can simplify it as shown above
and then enter it in. It is probably best to simplify it first, because
in some cases the numbers can get rather large, and it would be cumbersome
to multiply all those numbers one by one.
This means there is only 1 draw out of the 56 found in part a that would contain 3 RED marbles.
In part a above, we looked at all possible draws. From that list we only want the ones that contain 1 RED and 2 WHITE marbles.
Let's see what the draw looks like: we would have to have 1 red and 2 white marbles to meet this condition:
1 RED 2 WHITE
First we need to find n and r:
Note how 1 draw is split into two parts - red and white. We can not combine them together because we need a particular number of each one. So we will figure out how many ways to get 1 RED and how many ways to get 2 WHITE, and using the counting principle, we will multiply these numbers together.
1 RED:
If you said n = 3 you are correct!!! There are a total of 3 RED marbles.
If r is the number of RED marbles we are drawing at a time, what do you think r is?
If you said r = 1, pat yourself on
the back!! 1 RED marble is drawn at a time.
2 WHITE:
If you said n = 5 you are correct!!! There are a total of 5 WHITE marbles.
If r is the number of WHITE marbles we are drawing at a time, what do you think r is?
If you said r = 2, pat yourself on
the back!! 2 WHITE marbles are drawn at a time.
Let's put those values into the combination formula and see what
we get:
*Eval. inside ( )
*Expand 3! until it gets to 2!
*Expand 5! until it gets to 3!
*Cancel out 2!'s and 3!'s
If you don't have a factorial key, you can simplify it as shown above
and then enter it in. It is probably best to simplify it first, because
in some cases the numbers can get rather large, and it would be cumbersome
to multiply all those numbers one by one.
This means there are 30 draws that would contain 1 RED and 2 WHITE
marbles.
In part a above, we looked at all possible draws. From that list we only want the ones that contain 2 RED and 1 WHITE marbles. Remember that we need a total of 3 marbles in the draw. Since we have to have 2 red, that leaves us needing 1 white to complete the draw of 3.
Let's see what the draw looks like: we would have to have 2 red and 1 white marbles to meet this condition:
2 RED 1 WHITE
First we need to find n and r:
Note how 1 draw is split into two parts - red and white. We can not combine them together because we need a particular number of each one. So we will figure out how many ways to get 2 RED and how many ways to get 1 WHITE, and using the counting principle, we will multiply these numbers together.
2 RED:
If you said n = 3 you are correct!!! There are a total of 3 RED marbles.
If r is the number of RED marbles we are drawing at a time, what do you think r is?
If you said r = 2, pat yourself on
the back!! 2 RED marble is drawn at a time.
1 WHITE:
If you said n = 5 you are correct!!! There are a total of 5 WHITE marbles.
If r is the number of WHITE marbles we are drawing at a time, what do you think r is?
If you said r = 1, pat yourself on the back!! 1 WHITE marble are drawn at a time.
Let's put those values into the combination formula and see what
we get:
*Eval. inside ( )
*Expand 3! until it gets to 2!
*Expand 5! until it gets to 4!
If you don't have a factorial key, you can simplify it as shown above
and then enter it in. It is probably best to simplify it first, because
in some cases the numbers can get rather large, and it would be cumbersome
to multiply all those numbers one by one.
This means there are 15 draws that would contain 2 RED and 1 WHITE
marbles.
Practice Problems
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1b: A teacher has 15 students and 5 are to be chosen to give demonstrations. How many different ways can the teacher choose the demonstrators given the following conditions.
1a. The order of the demonstrators is important?
(answer/discussion to 1a)1b. The order of the demonstrators is not important?
(answer/discussion to 1b)
Practice Problems 2a - 2c: A teacher has 14 freshmen, 15 sophomores, 8 juniors and 10 seniors and 8 names will be drawn from a hat to go on a trip. How many different draws of student names can the teacher have given the following conditions.
2a. How many different draws of 8 names are there overall?
(answer/discussion to 2a)2b. How many different draws of 8 names would contain only juniors?
(answer/discussion to 2b)2c. How many different draws of 8 names would contain exactly 4 juniors and 4 seniors?
(answer/discussion to 2c)
Need Extra Help on these Topics?
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on May 20, 2011 by Kim Seward.
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