Hey there! Sign in to join this conversationNew here? Join for free

In how many different ways is it possible to multiply 5 variables? Watch

    • Thread Starter
    Offline

    11
    ReputationRep:
    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?
    Offline

    22
    ReputationRep:
    You have any product being of the form A^aB^bC^cD^dE^e, 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/
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    You have any product being of the form A^aB^bC^cD^dE^e, 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/
    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?
    Offline

    13
    ReputationRep:
    (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...
    Offline

    22
    ReputationRep:
    (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!.
    Offline

    18
    ReputationRep:
    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.
    Offline

    17
    ReputationRep:
    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.
    Offline

    17
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    You just want permutations then. That's just 5!.
    (Original post by TheMindGarage)
    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.
    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?
    Offline

    18
    ReputationRep:
    (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.
    Offline

    17
    ReputationRep:
    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!).
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by TheMindGarage)
    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.
    (Original post by DFranklin)
    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!).

    Thanks guys
    Offline

    4
    ReputationRep:
    There are 5! ways of multiplying 5 numbers
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.