Results 1 to 2 of 2

Math Help - Help

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    9

    Exclamation Help

    For the first number, there are 10 possible choices.

    For the second number, there are 9 possible remaining choices.

    For the third number, there are 8 possible remaining choices.


    Therefore, for selecting three numbers,

    . . there are: 10 \times 9 \times 8 \:=\:720 possible "combinations".


    this is the email i got, but is 720 the answer for all of the "combinations", or just the ones begining with 1??

    I AM CONFUSED
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    If you were to have the set, let's say S, containing all the integers 0-9, then there are 10 possible elements from which to choose. Now upon picking the second number there are 9 elements from which to choose. This assumes that there are no repeats. And finally as you said, there are 8 elements to from which to choose the third time picking. This shows that there are 720 distinct combinations or permutations of choosing three elements from this set.

    They don't have to start with 1.

    2,1,3 2,1,4... 2,1,9
    2,3,1, 2,3,4... 2,3,9
    etc.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum