The order of the demonstrators is important?
If you said a permutation problem, give yourself a pat on the
back..
First we need to find n and r :
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:
*Eval. inside ( )
*Expand 15! until it gets to 10!
*Cancel out 10!'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 360360 different ways the teacher can set
up the demonstrators with regard to order.
The order of the demonstrators is not important?
If you said a combination problem, give yourself a pat
on the back.
First we need to find n and r :
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:
*Eval. inside ( )
*Expand 15! until it gets to 10!
which is the larger ! in the den.
*Cancel out 10!'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 3003 different ways the teacher can set
up the demonstrators without regard to order.
How many different draws of 8 names are there overall?
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 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:
*Eval. inside ( )
*Expand 47! until it gets to 39!
which is the larger ! in the den.
*Cancel out 39!'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 314,457,495 different draws.
How many different draws of 8 names would 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 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:
*Eval. inside ( )
*Cancel out 8!'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 that would contain only juniors.
How many different draws of 8 names would contain exactly 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:
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 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 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:
*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 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.