math question, need help

[lendaddy]

I have forgotten how to figure this:

If you have a group of 39 numbers, how many unique seven number combinations are there within? And how do you figure it out?

[M.D. Holloway]

From Dr Math...

Here is a way to get the formula you asked for. Let's say there are N in all (so your question sets N equal to 7). List the combinations in increasing order, but to make things a little clearer, I will put an X every time you are finished listing all of one kind of number and go on to the next. For example, instead of

1 1 2 3 5 5 5

write

1 1 X 2 X 3 X X 5 5 5 X .

Notice there are 2 X's after the 3, or rather that there are
zero 4's between the third and fourth X. So now the list of
combinations begins like this:

1 1 1 1 1 1 1 X X X X X

1 1 1 1 1 1 X 2 X X X X

1 1 1 1 1 1 X X 3 X X X

1 1 1 1 1 1 X X X 4 X X

1 1 1 1 1 1 X X X X 5 X

1 1 1 1 1 1 X X X X X 6

1 1 1 1 1 X 2 2 X X X X

1 1 1 1 1 X 2 X 3 X X X

and so on. The useful thing about writing it this way is that
you can tell what the combination is merely by knowing the
positions of the five X's. In other words, there is a 1 to 1
correspondence between combinations of N dice and ways of
choosing 5 places to put an X out of a total of N+5 positions.
(Do you see why there are N+5 positions? In the above examples,
7 of the positions are filled by numbers and 5 get X's .) Now the
problem is easily solved: it is the binomial coefficient "N+5
choose 5", which is equal to:

(N+5) (N+4) (N+3) (N+2) (N+1) / 120 .

So for N=1 you get 6, for N=2 you get 21, for N=3 you get 56,
and so on.

[Mule]

Len, you been slappin' the cratch too much (or not enough).

[lendaddy]

Is there an answer in there and I missed it? It looks like a more difficult question than I started with

[Aerkuld]

Len - Do you mean that you have a set of 39 unique numbers (1 to 39), or that you have 39 numbers with multiples of the same number in the set (1,3,5,7,7,4,9,8,7,3,1,4,2,3,3,etc...)?

[lendaddy]

The numbers 1 through 39. Think of it as a lotto or keno set.

[Aerkuld]

Had to edit my answer when I read the post again, but with a set size = n, the combination size = k then:

Number of permutations (P)= n!/k!(n-k)!

Which is what jriera said below...

Where the ! denotes 'Factorial' which is the product of the number and all positive intergers below that number. So for 39! that would be 39x38x37xetc....x3x2x1

[jriera]

39!/(32!*7!) or 15.380.937 'unique' combinations (no repetitions)

The formula will be
n!
n_C_k = ---------- where n= 39 and k= 7 so
k!(n - k)!


39!
39_C_7 = ----------
7!(39 - 7)!

[Tim L]

39*38*37*36*35*34*33 if you cant use the number over like 7,7,7,7,7,7,7 if you can it's 39**7

[cmccuist]

L-Daddy, try this.



where:

r is the size of each permutation,
n is the size of the set from which elements are permuted, and
! is the factorial operator.
For your example, n = 39 and r = 7.

So it would be 39!/(39-7)! = 39!/32!=77,519,922,478

If you can't use the numbers over again it's that mess divided by 7!

See jriera's post.

[jriera]

Thanks Craig, I was not able to find a nice jpeg with the formula, I try to write it 'old school' but I failed.

[masraum]

Originally Posted by lendaddy

I have forgotten how to figure this:

If you have a group of 39 numbers, how many unique seven number combinations are there within? And how do you figure it out?

Wish you had never asked now, don't you. Yeah, it's kind of a pain in the butt to figure out.

[lendaddy]

Thanks guys, you rock!

To answer the commo0n question, no you cannot use a number twice in a combination. So am I correct that Jordi (and others) had it correctly at 15,380,937?

[lendaddy]

Originally Posted by cmccuist

L-Daddy, try this.



where:

r is the size of each permutation,
n is the size of the set from which elements are permuted, and
! is the factorial operator.
For your example, n = 39 and r = 7.

So it would be 39!/(39-7)! = 39!/32!=77,519,922,478

If you can't use the numbers over again it's that mess divided by 7!

See jriera's post.

hmmm, I divided 77,519,922,478 by seven and got a different number than Jordi's. Now I'm confused again

[jriera]

because is not divided by 7 but rather by 7! (7 factorial or 5.040)
77519922478/5040=15380937

Permutations and Combinations are fun!!!

[lendaddy]

Originally Posted by jriera

because is not divided by 7 but rather by 7! (7 factorial or 5.040)
77519922478/5040=15380937

Permutations and Combinations are fun!!!

Ahh, very cool. Thanks,

[Porsche_monkey]

I believe the answer is expressed as 39 choose 7, and as cmcuist said:

Now let's suppose we have 10 letters and want to make groupings of 4 letters. It's harder to list all those permutations. To find the number of four-letter permutations that we can make from 10 letters without repeated letters (10_P_4), we'd like to have a formula because there are 5040 such permutations and we don't want to write them all out!

For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. This is part of a factorial (see note).

To arrive at 10 x 9 x 8 x 7, we need to divide 10 factorial (10 because there are ten objects) by (10-4) factorial (subtracting from the total number of objects from which we're choosing the number of objects in each permutation). You can see below that we can divide the numerator by 6 x 5 x 4 x 3 x 2 x 1:


10! 10! 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
10_P_4 = ------- = ---- = --------------------------------------
(10 - 4)! 6! 6 x 5 x 4 x 3 x 2 x 1

= 10 x 9 x 8 x 7 = 5040

From this we can see that the more general formula for finding the number of permutations

[snowman]

DUH!! I knew that.