The digits 3,4,5,6,7, and 8 are randomly arranged to form a
three-digit number. (Digits are not repeated.) Find the
probability that the number is even and greater than 800.
The solution above is incorrect. He did not choose the MOST
RESTRICTIVE digit first, and got 12 ways for the numerator,
which is incorrect. Also he did not find the probability.
Here's the correct solution:
We find the number of even three-digit numbers greater than
800 with all different digits taken from 3,4,5,6,7,8
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 2 ways, as 4 or 6.
[We can't choose 8 because we must save it for the first
digit, since the number must be greater that 800].
We can then choose the first digit 1 way, as 8.
We can then choose the middle digit as any of the remaining 4 digits.
So that's 2∙1∙4 = 8 ways.
834, 836, 846, 854, 856, 864, 874, 876
So the numerator of the desired probability is 8.
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∙5∙4 = 120.
So the denominator of the desired probability is 120.
Answer is 8 ways out of 120 or 8/120 which reduces to 1/15.
Edwin