Permutation and Combination Problem

[sukhman]

A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2,4 and 5, which can be repeated. If the product of the digits in the area code must be even, how many different codes can be created? Answer = 3 × 3 × 3 - 1 = 26.

[Brent@GMATPrepNow]

A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2,4 and 5, which can be repeated. If the product of the digits in the area code must be even, how many different codes can be created?

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We know that the product of 3 digits is either odd or even.
So, we'll use the following fact: # of codes with EVEN product = total # of codes - # of codes with ODD product

total # of codes
Take the task of "building area codes and break it into stages.
Stage 1: Select 1st digit
We can choose a 2, 4 or 5, so we can complete this stage in 3 ways
Stage 2: Select 2nd digit
We can choose a 2, 4 or 5, so we can complete this stage in 3 ways
Stage 3: Select 3rd digit
We can complete this stage in 3 ways

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus build an area code) in (3)(3)(3) ways (= 27 area codes)

For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

# of codes with ODD product
If the product of 3 digits is odd, then EVERY digit must be odd.
So, the area code with an odd is 555.
So, there's only 1 area code with an ODD product.

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# of codes with EVEN product = 27 - 1 = 26

Cheers,
Brent

[Java_85]

27 because: 3*3*3 - 1 (which is 555) = 27