So i have 6 different symbols! I need to make as many combinations of 4 of the 6 symbols as possible without using the same symbol more than once in a combination. How many possible combinations are there than meet that criteria and i would appreciate a formula if thats ok too. Thanks.

2. 6*5*4*3

3. Is the * a multiplication sign?

4. If it is a multiplication sign then that would mean 360 combinations of 4! But i cant have the same symbol used twice in a combination.

5. Originally Posted by GOST
If it is a multiplication sign then that would mean 360 combinations of 4! But i cant have the same symbol used twice in a combination.
Is that an exclamation or 4 factorial?

Yes, that is a multiplication symbol.

Also, there are 6 choices the first time, 5 the next since we removed the item we already used, then 4 because 2 items are removed to avoid repetition, ....

6. exclamation sorry. i havent used this forum before so im not used to the math lingo :/
Yes yes, but for example lets say i have letters A B C D E F
I want to make as many combinations of four as possible out of them;
A B E D
A B D F
..so on..
but i cant have any repeat combinations and there cannot be any A A C E stuff going on either. No letter used twice.

7. 6 5 4 3
_ _ _ _

1st spot 6 choices, 2nd spot 5, 3rd 4, and 4th 3.

6*5*4*3 = 360

Didn't see you had the letters A-F.

8. Thank you sooo much!!!

9. I edited my original post.

I didn't notice you had letters A-F.

10. Wow it just seems there are so many choices ya know! Thanks for your help btw, lol ive been annoying but im visual and the layout helped

11. Just make spots for your choices when you do these problems.

12. Will do thanks!!!

13. Ok apperantly there is a missing piece to the puzzle. When trying to find all the possible combinations i came across multiple duplicates in combination. The spots are a great help yet they determine the total number of possible "orders" not combinations. For example when using the spots shown above i get this
6 5 4 3
_ _ _ _ This gives me 6 choices for spot 1 and 5 for spot 2 and so on so that the same letter isnt used over again in the sequence, BUT this is what happens:

A B C D A C B D
_ _ _ _ _ _ _ _ See. Although the sequence is different they are still the same combination :/ I need a way to determine only the possible combinations. Thank you.
_ _ _ _

14. nCr 6 choose 4

$\displaystyle \displaystyle \frac{n!}{r!(n-r)!}=\frac{6!}{4!(6-4)!}=\frac{6*5}{2!}=15$