Tom, Bill, Robert, Roger, and Terry are standing in a row for a group photo.

BTGModeratorVI wrote: ↑

Sun Jul 26, 2020 6:38 am

Tom, Bill, Robert, Roger, and Terry are standing in a row for a group photo. In how many different orders can the five men stand if Tom refuses to stand next to Roger?

A. 48
B. 64
C. 72
D. 96
E. 120

Answer: C
Source: Veritas Prep

No. of ways five men can stand to get photographed = 5! = 120;

Now in 120 ways, there are ways in which Tom and Robert stand next to each other and we must exclude those ways.

Let's consider Tom and Robert as one person, so the no. of ways four men can stand to get photographed = 4! = 24. We must double these ways as Tom and Robert can swap their positions, while they still are next to each other.

Total no. unrequited ways = 2*24 = 48

Thus, the required ways = 120 – 48 = 72

Correct answer: C

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Sun Jul 26, 2020 6:38 am

Tom, Bill, Robert, Roger, and Terry are standing in a row for a group photo. In how many different orders can the five men stand if Tom refuses to stand next to Roger?

A. 48
B. 64
C. 72
D. 96
E. 120

Answer: C
Source: Veritas Prep

Take the task of arranging the 5 men and break it into stages.

Stage 1: Arrange Bill, Robert, and Terry in a row
There are 3 people, so we can arrange them in 3! ways.

Now that we've arranged 3 men, we'll place a potential standing space on each side of these 3 men.
For example: ___ Terry ___ Robert ___ Bill ___
Notice that, when we place the 2 remaining men (Tom and Roger), in the 4 available spaces, we will be guaranteed that they are not next to each other.

Stage 2: Select a place for Tom to stand
There are 4 spaces available,, so we can complete this stage in 4 ways.

Stage 3: Select a place for Roger to stand
There are 3 remaining spaces, so we can complete this stage in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange all 5 men) in (3!)(4)(3) ways (= 72 ways)

Answer: C

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

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent