If all of the telephone extensions in a certain company must be even

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BTGModeratorVI: Legendary Member; Posts: 1223; Joined: Sat Feb 15, 2020 2:23 pm; Followed by:1 members

If all of the telephone extensions in a certain company must be even

by BTGModeratorVI » Sat Aug 08, 2020 7:04 am

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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

Answer: C
Source: Official guide

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Source: Beat The GMAT — Problem Solving | Sat Aug 08, 2020 7:04 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

Re: If all of the telephone extensions in a certain company must be even

by Brent@GMATPrepNow » Mon Aug 10, 2020 4:45 am

BTGModeratorVI wrote: ↑
Sat Aug 08, 2020 7:04 am
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

Answer: C
Source: Official guide

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: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this video: https://www.gmatprepnow.com/module/gmat- ... /video/775

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Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

Re: If all of the telephone extensions in a certain company must be even

by Brent@GMATPrepNow » Mon Aug 10, 2020 4:47 am

BTGModeratorVI wrote: ↑
Sat Aug 08, 2020 7:04 am
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

Answer: C
Source: Official guide

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

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

Re: If all of the telephone extensions in a certain company must be even

by Scott@TargetTestPrep » Wed Aug 12, 2020 2:20 pm

BTGModeratorVI wrote: ↑
Sat Aug 08, 2020 7:04 am
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

Answer: C
Source: Official guide

Solution:

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

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