math help

[momto3]

My daughter has officially surpassed my math knowledge. She was sick last week and her make up homework involves permutations and combinations. I understand the difference between the two, but have no idea how to solve the problems without a calculator.

An example of one of her questions is C(8, 2). Would someone please explain how to solve this?

Thanks from this math deficient mom!

[taterguy]

1. Take her to the OG and ask a waitress for help

You're asking the Lounge for math help?

[momto3]

9. Good question.

I may not post here a lot, but do read every day. And despite appearances, I happen to know that there are some very smart people here. Plus, at this point, the waitress may know more than I.

[rug]

2. 42

[dsc]

3. Ok here goes

The formula for permutations is as follows P(n,r) = P!/(n-r)! where 5! means 5*4*3*2*1 and 6! means 6*5*4*3*2*1 and so on. 0! is defined as 1. So P(8,2) is 8!/(8-2)! which is 8!/6! which is 8*7*6*5*4*3*2*1/6*5*4*3*2*1 which winds up being 8*7 = 56. C(n,r) is the same formula as permutations but you have to divide my r! to account for the fact that order doesn't matter. So C(8,2) is 8!/(8-2)!2! which is 8*7*6*5*4*3*2*1/6*5*4*3*2*1*2*1 or 56/2= 28.

Now if you don't have a calculator the easy way to do permutations is to think of having r picks from n items with no repetition. So P(8,2) would work like this. 8 choices for the first item times 7 choices for the second = 56. p(10,4) would be 10 choices for first item, 9 for second, 8 for third, and 7 for last. which is 5040. For combinations you can do the above and divide by r! (sorry no real shortcuts there). Hope this helped.

edit to correct typo.

[SwissTony]

5. Typo: C(8,2) is 8!/(8-2)!2! (=8!/(6!2!)) not 8!/(8-6)!2! (=8!/(2!2!))

No need to write out all the factorials (try it with C(100,2)).

8! = 8x7x6!

So C(8,2) = 8x7x6!/(6!x2!)

The 6!s cancel so you get C(8,2) = (8x7)/(2x1) = 56/2 =28

[dsc]

7. fair enough

If I were the kids teacher and the kid had the set up and formula I could give two figs about the right answer. On a graphing calc anyone can do the right answer if they have the set up right.

[momto3]

11. That's the problem.

I could do this on a calculator, but she is not allowed to use one in class. Thanks for you help!

[momto3]

10. Thank you!

I knew there had to be a simpler way to do this without having to write out all of the factorials. She is not aloud to use a calculator.

[JVS]

4. You're probably going to need a calculator for combinatorics.

First of all, read the math book. It should explain notation and how to calculate some things. The two big concepts should be permutations P, and combinations C. This might help supplement it. http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Combinatorics is essentially counting very large numbers of possibilities.

C(8,2)= 8!/(2!*(8-2)!)

[SwissTony]

6. IMO, students being introduced to permutations and combinations shouldn't be exposed

to examples that can't be calculated by hand.

[momto3]

13. I agree. eom

[retread]

8. If your daughter's class is being introduced to permutations and combinations are calculators even

allowed? If so, why?

[momto3]

12. She is not allowed to use a calculator.

This was part of my problem. I did not know how to tell her to do this without her having to calculate many factorials.

It has been too long since I did any of this in school. And, I know that I did not do this in 6th grade.