How many different ways are there to arrange . . .

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 am; Thanked: 1 times; Followed by:5 members

How many different ways are there to arrange . . .

by M7MBA » Sun Dec 10, 2017 10:21 am

How many different ways are there to arrange a group of 3 adults and 4 children in 7 seats if adults must have the first, third, and 7 seats?

A. 12
B. 144
C. 288
D. 1,400
E. 5,040

The OA is B.

Can any expert give me a detailed explanation of how can I solve this PS question? Thanks.

Quote

Source: Beat The GMAT — Problem Solving | Sun Dec 10, 2017 10:21 am

EconomistGMATTutor: GMAT Instructor; Posts: 555; Joined: Wed Oct 04, 2017 4:18 pm; Thanked: 180 times; Followed by:12 members

by EconomistGMATTutor » Sun Dec 10, 2017 11:36 am

Hello M7MBA.

We have to arrange 3 adults and 4 children in 7 seats in the following form: A C A C C C A.

For the first place, we have 3 options.
For the second place, we have 4 options.
For the third place, we have 2 options.
For the fourth place, we have 3 options.
For the fifth place, we have 2 options.
For the sixth place, we have 1 option.
For the seventh place, we have 1 option.

So, the total of options to arrange the 7 people is equal to 3*4*2*3*2*1*1 = 144.

So, the correct answer is B.

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.

GMAT Prep From The Economist
We offer 70+ point score improvement money back guarantee.
Our average student improves 98 points.

Quote

GMATWisdom: Master | Next Rank: 500 Posts; Posts: 100; Joined: Wed Nov 29, 2017 4:38 pm; Thanked: 14 times

by GMATWisdom » Sun Dec 10, 2017 12:42 pm

M7MBA wrote:How many different ways are there to arrange a group of 3 adults and 4 children in 7 seats if adults must have the first, third, and 7 seats?

A. 12
B. 144
C. 288
D. 1,400
E. 5,040

The OA is B.

Can any expert give me a detailed explanation of how can I solve this PS question? Thanks.

three adults can be placed in three seats in 3! ways = 3*2*!=6
four children can be placed in four seats in 4! ways =4*3*2*1=24
Hence total number of ways = 6*24= 144
So B is the correct option.

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 » Fri Sep 27, 2019 7:36 am

M7MBA wrote:How many different ways are there to arrange a group of 3 adults and 4 children in 7 seats if adults must have the first, third, and 7 seats?

A. 12
B. 144
C. 288
D. 1,400
E. 5,040

The OA is B.

Can any expert give me a detailed explanation of how can I solve this PS question? Thanks.

The 3 adults can be arranged in 3! = 6, and the 4 children can be arranged in 4! = 24 ways, so the total number of arrangements is 6 x 24 = 144.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]