If you have 50 numbers and choose 6 random ones, how many combinations are there?

Answer 1

50C6=15,890,700
yes im the first one,
yahooo!!!!!!!!!!

the word "combination" has a specific meaning, that means the arrangement matter. ex 1,2,3,4,5,6 and 1,3,2,4,5,6 are two combinations

if the arrangement doesnt matter like in a lotto, u use the term "permutation" then the answer is 50P6 = 1.144 x 10^10

be careful on how u use the word combination

im the first one and i explained the answer thoroughly, pick me as the best answer, lol

Answer 2

The answer depends a lot on how you choose the numbers and what you mean by a combination.

If you can use the same number over, there are 50 choices for each selection so there are 50^6 = 15,625,000,000 possibilities.

If you are drawing the numbers out of a hat so that one you draw a number you can't get it again, there are 50 choices for the first numbe, 49 for the 2nd, etc. giving 50*49*48*47*46*45 = 11441304000 possibilities.

It also matters whether you care about the order of the numbers. If there is no sequence, meaning the combination 42, 27, 33 is considered the same as 42, 33, 27 then there are less possibilities than otherwise. Since 6 numbers can be arranged in 6! = 720 ways, you divide the number of possibilites above by 720. For the numbers drawn from a hat, this gives 15,890,700
possibilities. For numbers where the same number can appear more than once, the simple 720 ways does not apply and things can get very complicated and probably beyond the scope of your question.

Answer 3

This is a very standard procedure in combinatorics:

To choose y things out of x, the formula (often "x choose y" because it is so common) is like this:
x! / (y!*(x-y)!)
This can be proven by using the formula for permutations. To choose, say, 4 permutations from 10 things, simply choose one (10 ways), the next (9 ways), and so on, to make the answer 10*9*8*7 = 5040 ways to choose those permutations. In general, it's the same as 10! / (10-4)! which equals (10*9*8*7*6*5*4*3*2*1) / (6*5*4*3*2*1) which equals 10*9*8*7.

Now that we know the formula for permutations, we also notice that permutations are the same thing as choosing combinations, then deciding the order! So, in this case, how many ways can you order 4 objects? 4! or 4*3*2*1 ways. So we see that #permutations = #combinations * 4!

So since we know that #permutations is equal to 10!/6!, we see that 10!/6! = #combinations * 4!. Divide by 4! and you get:
10!/6!4! = # combinations!

So we see that the general formula for choosing y things out of x things is x! / (y! * (x-y)!)

The answer to your question, by the way, is 50 choose 6 = 15,890,700.

Answer 4

6 * 50 = 300 combinations

Answer 5

in addition to needing to know if the numbers are replaced, you also have to know if order matters or not. For example, is 123456 the same as 132456. i think if order matters the answer is 50*49*48*47*46*45*44 and no replacement

if order doesnt matter and no replacement, 50c6 is the right answer

if order doesnt matter and there is replacment it is 50^6

if order does matter but there is replacement then I think its 50p6 but i am not sure of this one.

Answer 6

11,441,304,000 = 50*49*48*47*46*45
Unless you put them back in the pile before picking the next one, then it would be 50 *50 *50 *50 *50 *50

Answer 7

assuming no numbers can be repeated

Combination Formula
nCr = (n!)/(r!(n - r)!)

50C6 = (50!)/(6!(50 - 6)!)
50C6 = (50!)/(5! * 44!)
50C6 = 95344200

ANS : 95344200

Answer 8

It depends on if you can or cannot use the same number twice. I just got finished with that kind of stuff and you need to know if the numbers are being "replaced".

Answer 9

50C6 = 50!/6!/44!

Answer 10

can u use the same # twice if not im workin on it

if they cant be used twice its 50x49x48x47
if they can its50^6