SOLUTION: How many numbers divisible by 5 and lying 5000 and 6000(both inclusive) can be formed from the digits 5,6,7,8 and 9?

Algebra ->  Permutations -> SOLUTION: How many numbers divisible by 5 and lying 5000 and 6000(both inclusive) can be formed from the digits 5,6,7,8 and 9?      Log On


   



Question 662231: How many numbers divisible by 5 and lying 5000 and 6000(both inclusive) can be formed from the digits 5,6,7,8 and 9?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I understand the problem to mean that
we can only use the digits 5, 6, 7, 8, and 9.

To be between 5000 and 6000, the number will have to start with 5,
because the only number 5000 and 6000 (both inclusive)
that does not start with 5 is 6000, which requires three zeros.

Numbers divisible by 5 must end in 5 or 10.
If we can only use the digits 5, 6, 7, 8, and 9,
a number divisible by 5 will have to end in 5.

That only allows us to choose the two middle digits.
Nothing is said about not repeating digits,
so there are 5 independent choices for each of the middle digits,
which makes 5%2A5=25 choices overall.