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Posted by on Tuesday, August 13, 2013 at 12:13am.

How many 5-digit odd numbers can be made using the digits 3, 5, 6, 7 and 9, if digits cannot be repeated?

  • math help - , Tuesday, August 13, 2013 at 12:17am

    There are only 4 ways to choose the final digit, since it must be an odd number. After that one is chosen, any unused digits may be selected in the other places. So, selecting the digits right-to-left, there are

    4*4*3*2*1 = 96 ways to choose the digits.

  • math help - , Tuesday, August 13, 2013 at 12:23am

    if the digits 1,3,5.7,9, what is the different

  • math help - , Tuesday, August 13, 2013 at 12:26am

    if all the digits are odd, then there are 5 choices for the last digit. So, the number of choices changes to

    5*4*3*2*1 = 120

    was that not immediately clear?

  • math help - , Tuesday, August 13, 2013 at 12:30am

    If digits cannot be repeated,how many three digit numbers can be made using the digits 3, 5, 7 and 9 ?

  • math help - , Tuesday, August 13, 2013 at 12:44am

    I know if four digit using the digit 3,5,7,and 9 , the answer is 4*3*2*1=24,
    but if three digit using the digit 3, 5,7, and 9, the answer is 3*2*1=6?

  • math help - , Tuesday, August 13, 2013 at 3:46am

    no; there are 4 choices for the 1st digit, then 3, then 2, so

    4*3*2 = 24

    Start with the greatest number of choices and work down. You don't have to end at 1.

    A 3-digit number using the digits 1,3,5,7,9 would have 5*4*3 = 60 ways.

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