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1. -RM-
2. You have any product being of the form , where a,b,c,d,e range from 0 to 5 and you need a + b + c + d + e = 5. This is a easy combinatorics problem using the 'stars and bars' method, which you can read more about here https://brilliant.org/wiki/integer-e...star-and-bars/
3. -RM-
4. (Original post by UCASLord)
Thanks for your response, but I think I've phrased my question badly.

Say there are three variables, A, B and C.

I could multiply them as follows:
ABC
BAC
CBA
CAB
BCA
ACB

So 6 ways in total.

Surely, it would be more for when there's 5 variables?
I'm not too sure, but I believe that you have to use factorials. So, for three variables, it would be 3! which is 3x2x1=6
Therefore, for 5 variables, it would be 5!, which is 5x4x3x2x1=120 ways of multiplying it out.
The reason I believe this is, is because you can have 5 variables in the first place (any from A-E), and then for the second place, you can only have four as you used one up, and three for next and so on...
5. (Original post by UCASLord)
Thanks for your response, but I think I've phrased my question badly.

Say there are three variables, A, B and C.

I could multiply them as follows:
ABC
BAC
CBA
CAB
BCA
ACB

So 6 ways in total.

Surely, it would be more for when there's 5 variables?
You just want permutations then. That's just 5!.
6. So all you want is the number of permutations (the number of ways to arrange 5 distinct things)? It's just 5 factorial or 120.
7. What you're really looking for is the number of permutations (as Zacken said) of a 5-letter sequence A,B,C,D,E, that is, the number of different ordered arrangements of these 5 letters. The way to count this is to note that you have 5 choices for which letter you start the sequence with, then 4 choices for the second letter (since you've already used one), then 3 for the next, 2 for the next, then 1 for the last. This gives a total of 5x4x3x2x1 (i.e. 5!) possible permutations. Clearly this generalises to n letters, where there are n! possible permutations.
8. (Original post by UCASLord)
In how many different ways is it possible to multiply 5 variables?

Take, for instance, A, B, C, D and E. I know I could manually write out the possible solutions, but is there a nice algebraic way of doing this? Could you please explain?
I think you need to explain what you mean.
9. -RM-
10. (Original post by UCASLord)
Thanks guys.

What about if a few of the variables were the same?

Say I had ABBBB, or BBBAA, how would I work out the number of times these could be uniquely multiplied out?
I'm not a combinatorics expert, but I'll do my best.

The first one should be straightforward. If you look at the A, there are 5 possible places the A can go. The rest must be filled with Bs.

For the second example, there are 5 possible places the first A can go, and 4 that the second can go in (you can't put two As in the same place!). However, the order in which the As are assigned doesn't matter (ABABB is the same as ABABB), so you have to halve this. So 5x4/2 = 10 possibilities.

Another way to do it is:

(total letters)! / [(number of As)! * (number of Bs)!] where "!" represents factorial. If you try your two examples, you should get 5 and 10.
11. More generally, the number of unique arrangements is found by (total letters)! divided by (number of A's)! * (number of B's)! * ... * (number of Z's)!

(when calculating the bit you divide by you can of course ignore any letters that don't occur at least twice).

e.g. ABRACADABRA has 5A's 2B's, 2R's 1C and 1D. The number of unique arrangements is 11! / (5! * 2! * 2!).
12. -RM-
13. There are 5! ways of multiplying 5 numbers

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