# Answer to Question #23071 in Statistics and Probability for gilbert ramos

Question #23071
how many three digit numbers can be formed from the set 0,1,2,...,9 if repetitions are not allowed and the number must be divisible by a)10 b)25.
1
2013-01-29T09:08:09-0500
a) is divided by 10
There can be:
& 1) only one digit on the third place = 0, because the number must devisible by 10 ;
& 2) 9 digits on the first place (all digits from the set except 0),
& 3)8 digits on the 2-nd place (all 10 digits except two we have already used for the first and third places)

So the total number of possibilities is (by multiplying rule) N = 1*9*8 = 72

b) is divided by 25
The number is divided by 25 if and only if last two digits are divided by 25.
There are 3 variances for the last two digits
& 1) 25 or 75. In that case we can put 7 different digits on the first place (all 10 except 0, 5, 7 or 2)
& 2) 00. In that case we can put 9 different digits on the second place (all ten except 0)

So the total number of possibilities is (due to addition and multiplication rules):
& N = 2*7+ 1*9 = 23

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