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