The Purplemath Forums

by **miss_duhh** on Thu Jan 08, 2009 9:01 pm

Okay, our teacher always gives a "little" help with whatever we're learning and then throws homework at us. I'm pretty sure I get the Permutation, but the Combination is so confusing! The problems seem so simple, but I just can't do it.
Just to check if I'm doing this right, is this correct?
12P6 = 665,280

And, I'm pretty sure I did that right.
But just changing the P to a C has confused me, because it's not possible for me to do it the was I am with combination.
This is the problem:

12C6

I have no clue what to do, because I've tried doing it like permutation with the factorial stuff. (Ex: 4!/2! = 4*3*2*1/2*1), but I'm really not sure. I keep kinding on websites that the answer is 924, but I don't know how they got that.

**Sponsor**

by **stapel_eliz** on Thu Jan 08, 2009 9:27 pm

The formula for "_mP_n" is given by:

. . . . . $\frac{m!}{n!}$

Using m = 12 and n = 6, this gives:

. . . . . $\frac{12!}{6!}\, =\, \frac{1\times 2 \times ... \times 6 \times 7 \times ... \times 12}{1\times 2 \times ... \times 6}\, =\, 7 \times 8 \times ... \times 12\, =\, 665,280$

The formula for "_mC_n" is given by:

. . . . . $\frac{m!}{n!\left(m\, -\, n\right)!}$

Using m = 12 and n = 6, you should get the result you've found elsewhere. If you get stuck, please reply showing your work and reasoning.

Thank you!

Eliz.

by **miss_duhh** on Thu Jan 08, 2009 9:44 pm

Thank you so much!
I get it now. :thumb:

I just have two more question, if you don't mind..
I'm confused between the difference in permutation and combination, and I'm confused what n and r stand for. (The two variables: nCr)
These are three of the word problems I have:

1) Maggie has 8 different colored beads. She will pick 6 beads to thread with a string for a necklace. How many different necklaces can she make in order of the beads matters?

2) How many five-digit password numbers can be created using the digits: 1, 3, 4, 6, 7, and 9? Assume you can use each digit only once.

3) George has 12 good friends but can only invite 4 friends to see a rock concert with him. In how many ways can he select 4 friends to invite?

by **stapel_eliz** on Thu Jan 08, 2009 10:07 pm

miss_duhh wrote:I'm confused between the difference in permutation and combination, and I'm confused what n and r stand for.

Combinations are just that: different ways of combining options. Permutations are combinations where the order matters.

If you have ten ingredients have you have to toss six of them into a pot all at once, you would be doing "ten, choose six", because the order in which you toss them doesn't matter. You need only pick the combination of six ingredients.

On the other hand, suppose you're doing a science experiment where you put six of ten chemicals into a beaker, and the order in which the chemicals are added affects the results. You would then be doing "ten, permute six", because the order in which you toss them is does matter.

The variables "n" and "r" correspond to the "ten" and the "six" in the examples above.

Note: The exercises you have posted imply that you are expected to be familiar with the above terminology, along with the related concepts and formulas, and also with more-advanced material. :shock:

Eliz.

by **miss_duhh** on Fri Jan 09, 2009 2:05 am

Thank you very much.
I understand now.

Note: The exercises you have posted imply that you are expected to be familiar with the above terminology, along with the related concepts and formulas, and also with more-advanced material.

I know.
I have no nice way to put it, but my teacher isn't all that bright. (Not only that, last year I had a teacher who didn't get past intervals :shock:

because she taught science and math. But mostly science.)
She always gives us work, but never teaches, and she expects us to know it, too.
While half of the other graders have her (which are mostly failing), all the lucky ones got the other teacher and are passing with flying colors.
I may have to take some tutoring, but that's okay.
Thanks much,
April.
(:

by **DAiv** on Fri Jan 09, 2009 8:40 am

stapel_eliz wrote:The formula for "_mP_n" is given by:

. . . . . $\frac{m!}{n!}$

Using m = 12 and n = 6, this gives:

. . . . . $\frac{12!}{6!}\, =\, \frac{1\times 2 \times ... \times 6 \times 7 \times ... \times 12}{1\times 2 \times ... \times 6}\, =\, 7 \times 8 \times ... \times 12\, =\, 665,280$

I think there's a mistake with this permutations formula.

It should be:

$_mP_n=\frac{m!}{(m-n)!}$

Using m = 12 and n = 6, this gives:

$\frac{12!}{(12-6)!}\, =\, \frac{1\times 2 \times ... \times 6 \times 7 \times ... \times 12}{1\times 2 \times ... \times 6}\, =\, 7 \times 8 \times ... \times 12\, =\, 665,280$

The reason it didn't show up is because in this example, (m-n) happens to equal n.

To illustrate this, if we were to select any 1 from 4, there should only be 4 possible permutations.

Using m = 4 and n = 1, this gives:

$\begin{eqnarray}<br />\frac{m!}{n!} &=& \frac{4!}{1!} &=& \frac{1\times 2 \times 3 \times 4}1 &=& 2 \times 3 \times 4 &=& 24 &\text{ (Incorrect)}\\<br />\frac{m!}{(m-n)!} &=& \frac{4!}{(4-1)!} &=& \frac{1\times 2 \times 3 \times 4}{1 \times 2 \times 3} &=&&& 4 & \text{ (Correct)}\\<br />\end{eqnarray}$

DAiv

by **stapel_eliz** on Fri Jan 09, 2009 12:53 pm

DAiv wrote:
stapel_eliz wrote:The formula for "_mP_n" is given by:

. . . . . $\frac{m!}{n!}$

I think there's a mistake with this permutations formula. It should be:

$_mP_n=\frac{m!}{(m-n)!}$

Oops! You're quite right! :oops:

I didn't catch the typo because the value works out the same in this case.

Thank you! :thumb:

Eliz.

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