Formula for Possible Cominations?

Question

I want to know how to calculate how many possible combinations of 6 symbols out of 39 symbols can be made
Each symbol can only be used once in each combination (no AAAABC or ABBBBC)
You can't have the same combination twice, just with a mixed up order (ABCDEF, BDEACF, and FEDACB would all count as one, or the same combination)
It would be great if you could prove to me how/why the formula is or number of combinations you come up with is correct
Thank you a lot

Answer 1

1) Let's first consider not the combination but the permutations without repetition (considering the order).

Suppose to have 39 numbered cards in your hands and shuffle them well.
If you get one card and you drop it on the table you have 39 different possibilities, obviously.

Now, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.

If you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.

Recursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order.

2) Now let's see what happend when the order of the selection matters as you asked in your question.
So let's make this consideration:

a) If you have one card (A) you don't mind about ordering problems: {A} = (A)

b) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;

c) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).

So we can consider the 3 cards (A, B, C) as:
( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.
You could also group (A, {B, C}) = ( {B, C}, A) or {A, C} and then swapping with B.

So with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.

If we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.
So the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's 4 * (3 * 2).

With 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's 5 * (4 * 3 * 2).

Finally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.

Now, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).

In other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.

And now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements.

So Just divide 39*38*37*36*35*34 by 6*5*4*3*2 and you'll have the result of the combinations.

So to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.

the number 6*5*4*3*2 could be wrote "6!" using the the factorial function.

The factorial function of positive integer n, is defined as the product of first n numbers:

n! = 1 * 2 * 3 * ... * n

and is a convention that the product of no numbers at all is 1.

0! = 1


The factorial is also defined recursively by the follow definition:

(n+1)! = (n+1) * n!


So if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:

n! / (n-k)!

so it's the product of k terms starting from n and descending until reached (n-k+1).

n*(n-1)*(n-2)*...*(n-k+1) = n! / (n-k)!


Instead of permutation, if we consider the combination (the order doesn't matter) without repetition the formula is given by the permutation formula divided by k!.

n! / ( k! (n-k)! )

This function is called also Binomial Coefficient.

So for your answer it could be easily calculated in this way:

39! / (6! * (39-6)!) = 39*38*37*36*35*34 / (2*3*4*5*6) = 37*19*17*13*7*3 = 3262623

:p

Answer 2

You've got 39 things taken 6 at a time, so it's C(39,6).

The formula for combinations of n things taken r at a time is:
C(n,r) = n! / [r! (n - r)!]

In your example, C(39,6) = 39! / [6! * 33!], which simplifies to 3262623.

As to the how or why this formula is correct, first think of permutations.
You have 39 choices for the first number, 38 for the second, 37 for the third, and so on to 34 for the sixth. The number of permutations would be:
39 * 38 * 37 * 36 * 35 * 34, or 2349088560.

The trouble is, this allows for the six symbols you've chosen to be in some other order, a no-no in the condition to your problem. So, to get around this, we divide the 2349088560 by the total number of permutations in which these six can be arranged. That number is:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

2349088560 / 720 = 3262623.

Answer 3

Just think of it logically..
you have 6 spots for symbols...
_ _ _ _ _ _

you have 39 choices for the first symbol

39 _ _ _ _ _

now since you cannot repeat a symbol again, for the second spot you have now only a choice of 38

39 38 _ _ _ _

you keep going till you run out of spots for symbols, and multiply your choices together.

39*38*37*36*35*34

but there are 6! ways that the combinations can be counted again and again. So you must divide this total by 6!

so (39*38*37*36*35*34)/6!

That is why the above formula someone else stated works.
nCr - "n choose r" - where n is the total number of symbols and r is the sample size you want

then the formula is n!/{(n-r)!*r!}

Answer 4

I think you can use the combination formula: 39C6. If your calculator is good enough, you can do it. But I'll do it for you:
39C6 = 39P6/6P6 = 39*38*37*36*35*34/6*5*4*3*2*1 = a big number.

Answer 5

Permutation

nPr = (n!)/((n - r)!)

39P6 = (39!)/((39 - 6)!)
39P6 = (39!)/(33!)
39P6 = 39 * 38 * 37 * 36 * 35 * 34
39P6 = 2349088560

ANS : 2349088560

Answer 6

i comprehend it doesnt remember besides the undeniable fact that it must be finished this way... N=no longer something accessible consequences: Blue N N blue pink N N pink Redblue Bluered theres 6 consequences with this one there are 3 variables N rd and Blue so the reply is the variable factorial with 6 it could be 6 factorial = 720 in case you in person-friendly words wanted pairs it could be (6 x 5)/2

Answer 7

There is an algebraic formula for this. Long ago forgotten.

Go buy a BRIDGE TALLY. That'll do it for you.

Answer 8

the for mula is (x)*(x-1)*(x-2)...*(x-5)

x=39

i think other answerers will explain why :D