College Algebra
Answer/Discussion to Practice Problems
Tutorial 57: Combinations
 Answer/Discussion
to 1a
The order of the demonstrators is important? |
Keeping in mind that order is important, would this be a permutation
or a combination?
If you said a permutation problem, give yourself a pat on the
back..
First we need to find n and r :
If n is the number of students we have
to choose from, what do you think n is
in this problem?
If you said n = 15 you are correct!!!
There are 15 students.
If r is the number of students chosen at
a time, what do you think r is?
If you said r = 5, pat yourself on
the back!! 5 students are chosen to give demonstrations. |
Let's put those values into the permutation formula and see what
we get: |
|
*n = 15, r = 5
*Eval. inside ( )
*Expand 15! until it gets to 10!
*Cancel out 10!'s
|
If you have a factorial key, you can put it in as 15!, divided by
10! and then press enter or =.
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 360360 different ways the teacher can set
up the demonstrators with regard to order. |
Answer/Discussion
to 1b
The order of the demonstrators is not important? |
Keeping in mind that order is NOT important, would this be a
permutation or a combination?
If you said a combination problem, give yourself a pat
on the back.
First we need to find n and r :
If n is the number of students we have
to choose from, what do you think n is
in this problem?
If you said n = 15 you are correct!!!
There are 15 students to chose from.
If r is the number of students chosen at
a time, what do you think r is?
If you said r = 5, pat yourself on
the back!! 5 students are chosen to give demonstrations. |
Let's put those values into the combination formula and see what
we get: |
|
*n = 15, r = 5
*Eval. inside ( )
*Expand 15! until it gets to 10!
which is the larger ! in the den.
*Cancel out 10!'s
|
If you have a factorial key, you can put it in as 15!, divided by
10!, divided by 5! and then press enter or =.
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 3003 different ways the teacher can set
up the demonstrators without regard to order. |
Answer/Discussion
to 2a
How many different draws of 8 names are there overall? |
This would be a combination problem, because a draw would be
a group of names without regard to order.
Note that there are no special conditions placed on the names that we
draw, so this is a straight forward combination problem.
First we need to find n and r :
If n is the number of names we have to
choose from, what do you think n is in
this problem?
If you said n = 47 you are correct!!!
There are 14 freshmen names, 15 sophomore names, 8 junior names,
and 10 senior names for a total of 47 names.
If r is the number of names we are drawing
at a time, what do you think r is?
If you said r = 8, pat yourself on
the back!! 8 names are drawn at a time. |
Let's put those values into the combination formula and see what
we get: |
|
*n = 47, r = 8
*Eval. inside ( )
*Expand 47! until it gets to 39!
which is the larger ! in the den.
*Cancel out 39!'s
|
If you have a factorial key, you can put it in as 47!, divided by
39!, divided by 8! and then press enter or =.
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 314,457,495 different draws. |
Answer/Discussion
to 2b
How many different draws of 8 names would contain only juniors? |
This is a combination problem with a twist. In part a we looked at all possible draws. From that list we only want the
ones that contain only juniors.
Let's see what the draw looks like: we would have to have 8
juniors to meet this condition:
8 JUNIORS
First we need to find n and r :
If n is the number of JUNIORS we have to
choose from, what do you think n is in
this problem?
If you said n = 8 you are correct!!!
There are a total of 8 JUNIORS.
If r is the number of JUNIORS we are drawing
at a time, what do you think r is?
If you said r = 8, pat yourself on
the back!! 8 JUNIORS are drawn at a time. |
Let's put those values into the combination formula and see what
we get: |
|
*n = 8, r = 8
*Eval. inside ( )
*Cancel out 8!'s
|
If you have a factorial key, you can put it in as 8!, divided by
0!, divided by 8! and then press enter or =.
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 that would contain only juniors. |
Answer/Discussion
to 2c
How many different draws of 8 names would contain exactly 4 juniors
and 4 seniors? |
Again we have a combination with a condition attached. Whenever
you have a condition, you want to illustrate to yourself what would be
in the group or draw. In this case, the only possible way this could
come out is if you have 4 juniors and 4 seniors :
Let's see what the draw looks like: we would have to have 4
juniors and 4 seniors to meet this condition:
4 JUNIORS 4 SENIORS
First we need to find n and r:
Together that would make up 1 draw. We are going to have to use
the counting principle to help us with this one. If you need a review
on the Fundamental Counting Principle, feel free to got to Tutorial
55: The Fundamental Counting Principle.
Note how 1 draw is split into two parts - juniors and seniors.
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 4 JUNIORS and how many ways to get 4 SENIORS, and using the counting
principle, we will multiply these numbers together.
4 JUNIORS:
If n is the number of JUNIORS we
have to choose from, what do you think n is in this problem?
If you said n = 8 you are correct!!!
There are a total of 8 JUNIORS.
If r is the number of JUNIORS we
are drawing at a time, what do you think r is?
If you said r = 4, pat yourself on
the back!! 4 JUNIORS ARE drawn at a time.
|
4 SENIORS:
If n is the number of SENIORS we
have to choose from, what do you think n is in this problem?
If you said n = 10 you are correct!!!
There are a total of 10 SENIORS.
If r is the number of SENIORS we
are drawing at a time, what do you think r is?
If you said r = 4, pat yourself on
the back!! 4 SENIORS are drawn at a time. |
|
Let's put those values into the combination formula and see what
we get: |
|
*JUNIORS: n =
8, r = 4
*SENIORS: n =
10, r = 4
*Eval. inside ( )
*Expand 8! until it gets to 4!
*Expand 10! until it gets to 6!
*Cancel out 4!'s and 6!'s
|
If you have a factorial key, you can put it in as 8!, times 10!,
divided by 4!, divided by 4!, divided by 6!, divided by 4! and then press
enter or =.
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 14,700 different draws that would contain
4 juniors and 4 seniors. |
Last revised on May 16, 2011 by Kim Seward.
All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.
|
|
