Tom, Bill, Robert, Roger, and Terry are standing in a row

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

Tom, Bill, Robert, Roger, and Terry are standing in a row

by BTGmoderatorLU » Tue Oct 30, 2018 2:50 pm

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Source: Veritas Prep

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

The OA is C.

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Source: Beat The GMAT — Problem Solving | Tue Oct 30, 2018 2:50 pm

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

by Brent@GMATPrepNow » Tue Oct 30, 2018 3:05 pm

BTGmoderatorLU wrote:Source: Veritas Prep

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

The OA is C.

One approach is to ignore the rule and determine the total number of ways to arrange all 5 people, and then subtract the number of arrangements that BREAK the rule. I'm sure someone will post that kind of solution shortly.
Here's another approach:

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 our free 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
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- 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
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- 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

Brent Hanneson - Creator of GMATPrepNow.com

Quote

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pm; Thanked: 59 times; Followed by:33 members

by fskilnik@GMATH » Tue Oct 30, 2018 5:09 pm

BTGmoderatorLU wrote:Source: Veritas Prep

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

$$A,B,C,D,E$$
$$?\,\, = \,\,\# \,\,{\rm{orders}}\,\,{\rm{with}}\,\,A,B\,\,{\rm{not}}\,\,{\rm{together}}$$
(Change names and create expressions to help you unblock your brain!)
$$?\,\, = \,\,{\rm{Total}} - \left( {\# \,\,{\rm{orders}}\,\,{\rm{with}}\,\,A,B\,\,{\rm{together}}} \right)$$
$$?\,\, = \,\,5! - 2! \cdot 4! = 4! \cdot \left( {5 - 2!} \right) = 24 \cdot 3 = 72$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

Quote

GMAT/MBA Expert

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

by Scott@TargetTestPrep » Wed Oct 31, 2018 5:23 pm

BTGmoderatorLU wrote:Source: Veritas Prep

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

We are given that 5 people must stand in a row for a photo and need to determine in how many different ways they can stand if Tom refuses to stand next to Roger.

We can create the following equation:

total # of arrangements = # ways Tom is next to Roger + # ways Tom is NOT next to Roger.

Thus:

total # of arrangements - # ways Tom is next to Roger = # ways Tom is NOT next to Roger.

Let's determine the total number of arrangements and the number of ways Tom is next to Roger.

Since there are 5 people, the total number of arrangements is 5! = 120.

The total number of arrangements with Tom next to Roger can be calculated as follows:

[Tom-Roger] - [Bill] - [Terry] - [Robert]

Since Tom must be with Roger, notice there are 4! ways to arrange the entire group. However, we must remember that Tom and Roger can be arranged in 2! ways since it could be [Tom-Roger] or [Roger-Tom].

Thus, the number of ways Tom is next to Roger is 4! x 2! = 24 x 2 = 48.

Finally, the number of ways Tom is NOT next to Roger is 120 - 48 = 72.

Answer: C

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