Problem Solving

knight247 wrote:How many even 4 digit numbers can be formed by using the digits 0-9 so that no two digits are repeated?
A 2296
B 2396
C 2444
D 2456
E 2486

For most counting questions we can break the task into stages and apply the Fundamental Counting Principle (FCP).

Stage 1: Select units digit
Stage 2: Select tens digit
Stage 3: Select hundreds digit
Stage 4: Select thousands digit

Two important things:
1) When you use the FCP, always begins with the most restrictive stage
2) To handle the problematic situation where having a zero in the thousands position is no good, let's first determine the total number of 4-digit integers that may or may not have a zero in this position, and then we'll subtract the number of 4-digit integers that have a zero in this position.

4-digit integers that may or may not have a zero in the thousands position
Stage 1: can be accomplished in 5 ways (even digits only)
Stage 2: can be accomplished in 9 ways (once we select a digit for stage 1, there are 9 digits to choose from)
Stage 3: can be accomplished in 8 ways
Stage 4: can be accomplished in 7 ways
Total number of 4-digit integers = 5x9x8x7 = 2520

4-digit integers that have a zero in the thousands position
We'll begin the most restrictive stage, which is stage 4.
Stage 4: can be accomplished in 1 ways (must be zero)
Stage 1: can be accomplished in 4 ways (even digit other than 0)
Stage 2: can be accomplished in 8 ways
Stage 3: can be accomplished in 7 ways
Total number of 4-digit integers with zero in thousands position = 1x4x8x7 = 224

Final answer = 2520 - 225 = 2296, so the answer is A

Cheers,
Brent