Possible numbers - Permutations

[suchoudh]

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?
(1) 20
(2) 150
(3) 225
(4) 300
(5) 320

OA 4

[Brent@GMATPrepNow]

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?
(1) 20
(2) 150
(3) 225
(4) 300
(5) 320
OA 4

First ignore the two 2's restriction and count the number of possibilities. Then subtract all numbers that break the two 2's rule.

Stage 1: Select 1st digit. This can be 2, 4, 6 or 8 (4 possibilities)
Stage 2: Select 2nd digit. This can be 1, 3, 5 or 7 (5 possibilities)
Stage 3: Select 3rd digit. This can be 2, 3, 5 or 7 (4 possibilities)
Stage 4: Select 4th digit. This can be 0, 3, 6 or 9 (4 possibilities)
So, ignoring the two 2's rule, there are 4x5x4x4 (320) numbers possible.

Now consider how many numbers break the rule.
To do this, we will assume that the 1st and the 3rd digits are 2
We we get:
Stage 1: Select 1st digit. Must be 2 (1 possibility)
Stage 2: Select 2nd digit. This can be 1, 3, 5 or 7 (5 possibilities)
Stage 3: Select 3rd digit. Must be 2 (1 possibility)
Stage 4: Select 4th digit. This can be 0, 3, 6 or 9 (4 possibilities)
Total number of numbers is 1x5x1x4 (20).

So, the answer is 320 - 20 = 300

[suchoudh]

I was using only 3,6,9 as the numbers which are divisible by 3. Hence arrived at 225.

Thanks for your detailed explanation Brent. Much appreciated.