Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to

This topic has expert replies
Moderator
Posts: 7112
Joined: Thu Sep 07, 2017 4:43 pm
Followed by:23 members

Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to

by BTGmoderatorDC » Sun Nov 27, 2022 11:57 pm

00:00

A

B

C

D

E

Global Stats

Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?
(A) 240
(B) 480
(C) 720
(D) 1440
(E) 3600

OA A

Source: Magoosh

GMAT/MBA Expert

GMAT Instructor
Posts: 16184
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to

by [email protected] » Mon Nov 28, 2022 7:13 am
BTGmoderatorDC wrote:
Sun Nov 27, 2022 11:57 pm
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?
(A) 240
(B) 480
(C) 720
(D) 1440
(E) 3600

OA A

Source: Magoosh
Take the task of seating the 7 children and break it into stages.

We’ll begin with the most restrictive stage.

Stage 1: Arrange children A, B and G
Since child A has to sit next to both B & G, we can conclude that child A must sit BETWEEN B and G
There are only 2 options: BAG and GAB
So, we can complete stage 1 in 2 ways

IMPORTANT: Once we've arranged A, B and G, we can "glue" them together to form a single entity. This will ensure that A is between B and G

Stage 2: Arrange the single entity and the four remaining children
There are 5 objects to arrange: C, D, E, F and the BAG/GAB entity.
We can arrange n different objects in n! ways
So, we can arrange the 5 objects in 5! ways (5! = 120)
So, we can complete stage 2 in 120 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 7 children) in (2)(120) ways (= 240 ways)

Answer: A
Brent Hanneson - Creator of GMATPrepNow.com

GMAT/MBA Expert

GMAT Instructor
Posts: 6800
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to

by [email protected] » Thu Dec 01, 2022 11:11 am
BTGmoderatorDC wrote:
Sun Nov 27, 2022 11:57 pm
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?
(A) 240
(B) 480
(C) 720
(D) 1440
(E) 3600

OA A

Source: Magoosh
In each of the hundreds (i.e., 1-99, 100s, 200s, etc.) of the first 1000 positive integers, except for the 700s, the digit 7 appears in 19 integers, either as the units digit or the tens digit or both. In the 700s, 7 appears in every integer; therefore, the total number of integers that has (at least) one digit of 7 is:

9 x 19 + 100 = 171 + 100 = 271

Answer: E

• Page 1 of 1