Question 1058258
The digits 2, 3, 4, 5, 6, and 7 are randomly arranged to form a 
three-digit number.  (Digits are not repeated.) Find the 
probability that the number is even and greater than 700. 
(Answer must be an integer or a simplified fraction.)
<pre>
We find the number of even three-digit numbers greater than
800 with all different digits taken from 2,3,4,5,6,7

The last digit is the MOST RESTRICTIVE, since it must be even.
so we choose it first:

We can choose the last digit any of 3 ways, as 2, 4, or 6.

We can then choose the first digit 1 way, as 7.

We can then choose the middle digit as any of the remaining 4 digits.

So that's 3&#8729;1&#8729;4 = 12 ways.

724, 726, 732, 734, 736, 742, 746, 752, 754, 756, 762, 764

So the numerator of the desired probability is 12.

The denominator is the number of permutations of 6 digits taken
3 at a time or 6P3 = 120. Or do it this way:

Choose the 1st digit 6 ways, the 2nd digit 5 ways, the 3rd digit 4 ways,
which is 6&#8729;5&#8729;4 = 120.

So the denominator of the desired probability is 120.

Answer is 12 ways out of 120 or 12/120 which reduces to 1/10.

Edwin</pre>