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If all of the telephone extensions in a certain company must

by BTGmoderatorLU » Thu Dec 06, 2018 2:49 pm

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Source: Princeton Review

If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of four-digit extensions that the company can have?

A. 4
B. 6
C. 12
D. 16
E. 24

The OA is C.
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by Brent@GMATPrepNow » Thu Dec 06, 2018 2:53 pm
BTGmoderatorLU wrote:Source: Princeton Review

If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of four-digit extensions that the company can have?

A. 4
B. 6
C. 12
D. 16
E. 24

The OA is C.
Take the task of arranging the 4 digits and break it into stages.

We'll begin with the most restrictive stage.

Stage 1: Select the digit in the units position
Since the 4-digit extension must be EVEN, the unit digit must be either 2 or 6
So, we can complete stage 1 in 2 ways

Stage 2: Select the tens digit
There are 3 remaining digits from which to choose, so we can complete this stage in 3 ways.

Stage 3: Select the hundreds digit
There are 2 digits remaining, so we can complete this stage in 2 ways.

Stage 4: Select the thousands digit
There 1 digit remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create an even extension) in (2)(3)(2)(1) ways (= 12 ways)

Answer: C

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by Brent@GMATPrepNow » Thu Dec 06, 2018 2:54 pm
BTGmoderatorLU wrote:Source: Princeton Review

If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of four-digit extensions that the company can have?

A. 4
B. 6
C. 12
D. 16
E. 24

The OA is C.
NOTE: Always SCAN the answer choices before performing any calculations.
Here, we notice that the answer choices are pretty small, so if we don't come up a full solution, we should consider LISTING AND COUNTING

We get the following extensions:
1236
1326
1362
1632
2136
2316
3126
3162
3216
3612
6132
6312
TOTAL = 12
Answer: C

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by Iakgmat » Thu Dec 06, 2018 6:37 pm
To use all 1/2/3/6 in a 4-digit number, this means no number can be used twice.

Thus if we consider _ _ _ _, the last digit must either be 2 or 6.

For the first three digits: the first digit we have 3 choices, 2 for the second, 1 for the third.

Thus, 3*2*1. We can do this twice (with 2 or 6 as ending, yielding 12.
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by ceilidh.erickson » Fri Dec 07, 2018 1:21 pm
Here is yet another solution:

Most of the time when we're arranging n things (people in a line, letter combinations, etc) with a diminishing pool (i.e. we can't use the same item twice), the number of arrangements is simply n!. If we could arrange these 4 numbers into extensions without any restrictions, we'd have 4 options for the 1st number, 3 for the 2nd, 2 for the 3rd, 1 for the 4th:
4*3*2*1 = 4! = 24.

These passcodes have to be even, though. Two of our digits are even (2 and 6) and two are odd (1 and 3). It thus stands to reason that HALF of the 24 combinations we came up with will end with an odd digit, and half with an even. If we only want the even results, we simply divide by 2:
24/2 = 12.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by Scott@TargetTestPrep » Sun Mar 24, 2019 5:08 pm
BTGmoderatorLU wrote:Source: Princeton Review

If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of four-digit extensions that the company can have?

A. 4
B. 6
C. 12
D. 16
E. 24

The OA is C.
Since the last digit of the extension must be even, the last digit can be either 6 or 2. If we let 6 be the last digit, then we have 3! = 6 options for the other 3 digits. If we let 2 be the last digit, we also have 6 options. Therefore, we have a total of 12 options.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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