Discrete mathematics
So, anyone here take discrete math at any point in their life and still remember anything about? I've ran into a problem that I'm not sure about, and my professor is terrible at teaching during class. He's fine one on one, but due my car circumstances I can't get in there before next class. Anywho, problem is this:
Show that if n > 1, then P(n,2) = n^2 - n
Not sure how to go about doing this, if I could just get some assistance that would be great. Not asking for the answer, just help. I would like to actually figure out how to do this so I'm not screwed come exam time
Thanks
-Biscuit
Show that if n > 1, then P(n,2) = n^2 - n
Not sure how to go about doing this, if I could just get some assistance that would be great. Not asking for the answer, just help. I would like to actually figure out how to do this so I'm not screwed come exam time
Thanks
-Biscuit
P(x, y) is the permutation function.
n!
-----
(n-k)!
if i recall correctly.
When your y is 2, think about it. All your actually doing is
n * (n - 1) since everything else cancels out.
For example P(10, 2) is
10!
----
(10-2)!
or
10x9x8x7x6x5x4x3x2x1
------------------------------
8x7x6x5x4x3x2x1
Which leaves you with 10 * 9 or just n * (n-1). So now we need to show that n^2 - n = n * (n-1). If you distribute n, you'll see that n * (n-1) = n^2 - n, and then by transitive property the two are equal. So therefore, if n > 1, P(N, 2) = n * (n-1) = n^2 - n.
I might be overstepping my bounds here a little by doing your homework, but I love combinatorics too much to stop :/
n!
-----
(n-k)!
if i recall correctly.
When your y is 2, think about it. All your actually doing is
n * (n - 1) since everything else cancels out.
For example P(10, 2) is
10!
----
(10-2)!
or
10x9x8x7x6x5x4x3x2x1
------------------------------
8x7x6x5x4x3x2x1
Which leaves you with 10 * 9 or just n * (n-1). So now we need to show that n^2 - n = n * (n-1). If you distribute n, you'll see that n * (n-1) = n^2 - n, and then by transitive property the two are equal. So therefore, if n > 1, P(N, 2) = n * (n-1) = n^2 - n.
I might be overstepping my bounds here a little by doing your homework, but I love combinatorics too much to stop :/
Last edited on
Well, P(n,r) = n!/(n-r)!
This is about where I'm at. Like I said, prof. is pretty bad at lecturing and the book for the course is maybe entirely useless. Sorry for the lack of info :/
This is about where I'm at. Like I said, prof. is pretty bad at lecturing and the book for the course is maybe entirely useless. Sorry for the lack of info :/
Sounds like you have to do a proof by induction.
EDIT: Or, you would if everything didn't cancel out so nicely.
EDIT: Or, you would if everything didn't cancel out so nicely.
Last edited on
Oh formulae. Just go with the definition and learn to count (for this particular problem)
Simply the factorial and then expand the quantity. Afterwards, a simple proof to show your algebraic solution equals what you are trying to prove.
I'll post a complete solution tommorow. I'm typing this on my phone and getting tired with this small keyboard.
I'll post a complete solution tommorow. I'm typing this on my phone and getting tired with this small keyboard.
n!/(n_r)!
N!/(n_2)!
Expand: N x (n_1) x (n_2)! / (n_2)!
Factorials cancel
n x (n_1) = n^2 _ n
The underscore is supose to be minus.
N!/(n_2)!
Expand: N x (n_1) x (n_2)! / (n_2)!
Factorials cancel
n x (n_1) = n^2 _ n
The underscore is supose to be minus.
You guys make me want to shoot myself for having stopped at an associates degree...
I took Discrete Mathematics at DeVry University for their Game and Simulation Programming degree, but sadly I don't remember much about it other than it was rather difficult.
Oh wow this is easier than it looked. This class as a whole so far is pretty simple, but I do have take discrete II next semester so that'll probably be harder.
Lol dont feel bad computergeek! You've been one of the most helpful people on this forum for me for various problems
Lol dont feel bad computergeek! You've been one of the most helpful people on this forum for me for various problems
I don't get why everyone posted the answer after I did, but glad we could all help anyways. Good luck with the course.
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