math help
posted by Tomas .
How many 5digit odd numbers can be made using the digits 3, 5, 6, 7 and 9, if digits cannot be repeated?

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 righttoleft, there are
4*4*3*2*1 = 96 ways to choose the digits. 
if the digits 1,3,5.7,9, what is the different

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

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