## Six children - A, B, C, D, E, and F

##### This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 48
Joined: 23 Apr 2013
Thanked: 6 times

### Six children - A, B, C, D, E, and F

by gmat_guy666 » Sun Sep 14, 2014 11:22 am
Six children - A, B, C, D, E, and F - are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A)60
B)180
C)240
D)360
E)720

OA

D

### GMAT/MBA Expert

GMAT Instructor
Posts: 15885
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1267 members
GMAT Score:770
by [email protected] » Sun Sep 14, 2014 11:33 am
gmat_guy666 wrote:Six children - A, B, C, D, E, and F - are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A)60
B)180
C)240
D)360
E)720
First, let's IGNORE the rule about child E seated somewhere to the left of child F.
When we ignore the rule, we can arrange the 6 children in 6! ways (= 720 ways)

Now recognize that, in HALF of those 720 arrangements, child E is seated somewhere to the LEFT of child F, and in the other HALF of those 720 arrangements, child E is seated somewhere to the RIGHT of child F.

So, the number of arrangements where child E is seated somewhere to the left of child F = (1/2)720 = 360

NOTE: This question is almost identical to this one: https://www.beatthegmat.com/counting-six ... tml#198825

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com

### GMAT/MBA Expert

GMAT Instructor
Posts: 15885
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1267 members
GMAT Score:770

### GMAT/MBA Expert

Elite Legendary Member
Posts: 10392
Joined: 23 Jun 2013
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:508 members
GMAT Score:800
by [email protected] » Sun Sep 14, 2014 1:10 pm
Hi gmat_guy666,

Brent's explanation is arguably the most efficient way to get to the correct answer. Here's a more drawn-out explanation for the correct answer:

We're given 6 spots:

_ _ _ _ _ _

And we have to put a child in each spot; E MUST be to the left of F though. Here's one possible way that this can occur:

E F _ _ _ _

In this option, any of the remaining 4 children can be placed in each of the 4 remaining spots, so we have....

E F (4)(3)(2)(1) = 24 options with E and F in the 1st and 2nd spots, respectively.

This pattern will occur over-and-over. For example....If E and F were in other places....

(4)E(3)(2)F(1) = 24 options with E and F in the 2nd and 5th spots, respectively.

So we really just need all of the possible placements for E and F, then we can multiply that result by 24...

While this is not necessarily the most efficient way to approach this task, the work isn't that hard....

E F _ _ _ _
E _ F _ _ _
E _ _ F _ _
E _ _ _ F _
E _ _ _ _ F

_ E F _ _ _
_ E _ F _ _
_ E _ _ F _
_ E _ _ _ F

_ _ E F _ _
_ _ E _ F _
_ _ E _ _ F

_ _ _ E F _
_ _ _ E _ F

_ _ _ _ E F

15 possible placements for E and F (given the restriction that E must be to the left of F).

15 x 24 = 360

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]

GMAT Instructor
Posts: 15533
Joined: 25 May 2010
Location: New York, NY
Thanked: 13060 times
Followed by:1901 members
GMAT Score:790
by GMATGuruNY » Sun Sep 14, 2014 4:20 pm
gmat_guy666 wrote:Six children - A, B, C, D, E, and F - are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A)60
B)180
C)240
D)360
E)720
One more approach:

There are no constraints on A, B, C, and D.
Number of options for A = 6. (Any of the 6 chairs.)
Number of options for B = 5. (Any of the 5 remaining chairs.)
Number of options for C = 4. (Any of the 4 remaining chairs.)
Number of options for D = 3. (Any of the 3 remaining chairs.)

Of the 2 remaining chairs, the one most to the left must be occupied by E, since E must sit to the left of F.
Number of options for E = 1. (The chair most to the left.)
Number of options for F = 1. (Only 1 chair remains.)

To combine the options above, we multiply:
6*5*4*3*1*1 = 360.

Mitch Hunt
Private Tutor for the GMAT and GRE
[email protected]

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
Student Review #1
Student Review #2
Student Review #3

Legendary Member
Posts: 1097
Joined: 10 May 2014
Location: New Delhi, India
Thanked: 205 times
Followed by:24 members
by GMATinsight » Sun Sep 14, 2014 9:01 pm
An Alternate methods is:

Select two places for B and E and there is only 1 way for them to sit on those chair, the chairs for B and E can be selected in 6C2 ways = 15

Now remaining 4 chairs can be occupied by other 4 people in 4! ways = 24 ways

Total ways : 6C2 x 4! = 15 x 24 = 360