SOLUTION: The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? Express youranswer

In order to have a success:

The tens digit must be a 3 or a 7 since it must be odd. 

Any number divisible by 4 is such that the 2-digit number 
formed by the last two digits is divisible by 4.  The 
only 2-digit numbers beginning with 3 which are divisible
by 4 are 32 and 36. The only 2-digit numbers beginning 
with 7 which are divisible by 4 are 72 and 76.  

Therefore,

there are only 4 types of numbers this can come out
and be successful, they are:

1.    _ _ _ 3 2

or

2.    _ _ _ 3 6

or

3.    _ _ _ 7 2

or

4.    _ _ _ 7 6

 
For any one of these four we can choose the first digit
any of the remaining 3 ways, the 2nd digit any of the
then remaining 2 ways, and there will be only one remaining
choice for the 3rd digit.  That's 3! for each of the 4 types,

So the numerator is 4*3!

The denominator is the number of ways of choosing ANY 5-digit
number.

We may choose the 1st digit any of 5 ways, the 2nd digit any
of of the 4 remaining digits, the 3rd digit any of the 3 
remaining digits, the 4th digit any of the 2 remaining digits
and the 5th is the 1 remaining digit.

So the denominator is 5*4*3*2*1 or 5!

Answer = 

Edwin