Malagasy Rain

I was studying permutations earlier this morning and I noticed 2 things concerning situations (the latter of which might be stated already).

Case 1:
When you have x things taken x-1 at a time (e. g. P(7,6) or P(9,8)), the total number of permutations will always be x!.

Ex:
Total permutations of 10 things taken 9 at a time.
P(10, 9) = 10!/(10-9)!
=10!/1!
=10!
=3628800
P(10, 9)=10!

Case 2 (the latter): When you have x things taken 1 at a time (e.g. P(5, 1) or P(2, 1)), the total number of permutations will always be x.

Ex:
Total permutations of 5 things taken 1 at a time.
P(5, 1) = 5!/(5-1)!
= 5!/4!
= 5*(4!/4!)
= 5*1
=5

P(5, 1) = 5

Pretty perceptive, huh?

nermal

Think it through. Let's say you and I are sitting across a table with x marbles on it.
You take 1 marble, leaving me x-1 marbles. First permutation.
You replace that marble, and take another, leaving me a different set of x-1 marbles.
Now, it's obvious that you will be able to pull exactly x different marbles.
So at this point, it should be obvious that you will, in doing so, leave me exactly x different permutations of x-1 marbles.

Yes, very perceptive.

Ed

pmurray

Case 1 is to be expected because once you have taken x-1 items, there's only one way to take the remaining one.

But yes, I think I know what you mean. It's amazing how the same answer drops out of verbal reasoning like the above as comes out of mechanically doing the equations. Truth has no loopholes (unlike certain religious texts I might mention).

Malagasy Rain

Concerning combinations, the total combinations of x things taken one at a time is always x.

Ex:
Combinations of 4 things taken 1 at a time.

C(4, 1) = 4!/1!(3!) = 4

P(x, 1) = C(x, 1) = C(x, x)

Anyone else notice that?

Principia

Try this. C(n, r) = n*(n-1)*(n-2)*...*(n-r+1) / [r * (r-1) * (r-2) *... *1] is always an integer, so long as n and r are non-negative integers, and n >= r. Why must that be?

EDIT: let me clarify. If I just gave you the following -- n*(n-1)*(n-2)*...*(n-r+1) / [r * (r-1) * (r-2) *... *1] how would you deduce with the given constraints on n and r that it would always result in an integer?

nermal

I need to get my text and look, but I'm betting you can factor all the r, r-1, r-2 etc out of the numerator and cancel leaving a 1 in the denominator. It's been a while since I've played with this.

Ed

Quantum Ninja

How about this:

The total combinations of x things taken n at a time is always equal to the total combinations of x things taken (x - n) at a time.

C(x, n) = C(x, x - n)

For example:
C(5, 2) = C (5, 3)

It's easy to see why this is true.

If I have five different colored marbles, and you take a combination of two from me, you leave me with a combination of three. For each combination of two that you take, there is a corresponding combination of three that I will be left with. Or if I have 100 marbles and you take combinations of three, for each combination you take, there will be a corresponding combination of 97 that I'm left with.

Actually, this is not true unless x=1.

P(x,1) = C(x,1) is true... you're good so far.

But C(x,x) is always equal to 1. P(x,1) and C(x,1) are always equal to x. Hence, your equality does not always hold.

Malagasy Rain

Damn. And I had my mathematical momentum going.