# Combination of n things taken k at a time?

• Jul 20th 2014, 04:20 PM
InceptSean
Combination of n things taken k at a time?
I understand how to do the normal permutations that are like

9!/(9-4)!

9x8x7x6x5!/5!

5!'s cancel, 9x8x7x6=3024 combinations, yeah yeah yeah....but I don't understand this other permutation thing for the life of me.

What's canceling, what's doing what? They don't explain, give explicit solution, thus how am I supposed to learn? The book doesn't even give explicit solution. Math always does this to me...anyways...:

http://i61.tinypic.com/6jqeiq.png
• Jul 20th 2014, 04:35 PM
romsek
Re: Combination of n things taken k at a time?
Not really sure what you're asking.

You might think about the Choose(n,k) function like this.

If order matters you have $n!$ permutations of $n$ items. That should be pretty clear.

Now when order doesn't matter you have permutations that are indistinguishable from one another.

A group of $n$ consists of the $k$ chosen, and the $n-k$ that aren't chosen.

For a given group of $k$ there are $k!$ ways those $k$ people can be arranged.

Similarly for the $n-k$ people not chosen, there are $(n-k)!$ ways they can be arranged.

So a given group of $k$ chosen people out of $n$ has $k! \times (n-k)!$ ways being arranged.

So if there are $n!$ total ways of being arranged and a choice of $k$ people can be arranged in $k! (n-k)!$ ways there are

$\dfrac {n!}{k!(n-k)!}$ ways to choose those $k$ people from the total $n$.

You did (a) just fine.

(b) should be pretty obvious. If 11 take part out of 23, how many don't take part? So how many ways can you choose those that don't take part out of 23?

Does this agree with your intuition?
• Jul 21st 2014, 09:25 AM
JeffM
Re: Combination of n things taken k at a time?
For a somewhat different explanation.

Let x = how many ways we can permute k distinct objects.

Let y = how many ways we can choose k distinct objects from n distinct objects.

Let z = how many ways we can choose k distinct objects from n distinct objects and then permute the chosen k objects.

Do you see that $z = xy?$ For each of the y ways we can choose the k distinct objects, there are x ways to permute them.

$z = xy \implies y = \dfrac{z}{x}.$ Basic algebra.

And we know $x = k!$

So $y = \dfrac{z}{k!}.$ Simple substitution.

But what is z?

I have {n - 0} = n ways to choose the first object.

I have {n - 1} ways to choose the second object, which gives me (n - 0)(n - 1) ways to choose the first and second objects.

Keeping on like that, I have

$\{n - 0\} * \{n - 1\} * ... \{n - (k - 1)\} = \dfrac{n!}{(n - k)!}$ ways to choose k distinct objects from n distinct objects and then permute the chosen k.

$y = \dfrac{z}{k!} = z * \dfrac{1}{k!} = \dfrac{n!}{(n - k)!} * \dfrac{1}{k!} = \dfrac{n!}{k! * (n - k)!}.$

Does that make sense?
• Jul 21st 2014, 04:35 PM
InceptSean
Re: Combination of n things taken k at a time?
I guess I didn't make my question specific enough. How did we arrive at 1,352,078?

23x22x21x20x19x18x17x16x15x14x13x12!/11!12! = 23x22x21x20x19x18x17x16x15x14x13/11!x1

Then, what?

I do 23x22x21x20x19x18x17x16x15x14x13/11 = 4.906420646E^12

You can see I am completely missing out on this concept. I probably messed up either by canceling out 12!'s when I wasn't supposed to or by diving by 11---I probably should have done something with the 11!, but I didn't. I randomly dropped the exclamation point, but I really don't know what to do, so that's probably why i dropped the exclamation point.

Hopefully my problem is explicit now. =[

p.s. the solution was a textbook solution, not a solution that I made, which is why I am questioning it.
• Jul 21st 2014, 06:32 PM
InceptSean
Re: Combination of n things taken k at a time?
I figured out I can just use the ti83 calculator for this....ughh.....lol. Problem solved, thank you everyone!