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

Redeem

Permutation - Combination (Counting) Question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Wed Sep 17, 2008 8:35 pm

Permutation - Combination (Counting) Question

by sukhpreet » Sun Nov 23, 2008 11:18 am
Q. How many ways can 2 students be seated in a row of 4 desks so there is always at least one empty desk between the students?
a) 2
b) 3
c) 4
d) 6
e) 12

correct ans: 6

Couldn't figure out how to tackle this question. Infact, the bigger problem has been that I usually make mistake with Probability and Permutation/Combination questions. Any suggestions about how can I strengthen my knowledge in these areas? Any useful resources? Thanks a bunch.
Top
Source: — Problem Solving |
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Wed Sep 17, 2008 8:35 pm

by sukhpreet » Sun Nov 23, 2008 1:38 pm
How should I solve this question to reach the correct answer? I came up with 4P3. 3 desks to choose from 4 (one being empty)
Top
User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

Re: Permutation - Combination (Counting) Question

by logitech » Sun Nov 23, 2008 2:02 pm
sukhpreet wrote:Q. How many ways can 2 students be seated in a row of 4 desks so there is always at least one empty desk between the students?
a) 2
b) 3
c) 4
d) 6
e) 12

correct ans: 6

Couldn't figure out how to tackle this question. Infact, the bigger problem has been that I usually make mistake with Probability and Permutation/Combination questions. Any suggestions about how can I strengthen my knowledge in these areas? Any useful resources? Thanks a bunch.

1XXX 2 has two places
X1XX 2 has only one place
XX1X 2 has only one place
XXX1 2 has two places

2+1+1+2 = 6
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Wed Oct 01, 2008 3:55 pm
Thanked: 6 times

by rmpaes » Sun Nov 23, 2008 2:14 pm
The easiest way to solve this problem is to first find how many ways 2 people can sit in 4 empty desks and then minus out the arrangements where they sit together.

When the first of 2 people first see the seats he sees 4 seats. If he sits down the next person will see 3 seats.
So the total possible arrangements is 4 x 3 = 12
or P(4,2) = 4!/2!

At least one seat between them means they don't sit together. So we find the arrangements where they sit together and minus this from the total arrangements.

So what we do is stick the two of them together. That means that they will take two seats. So when the couple decides to sit down they see 3 slots instead of 4 since they take two seats. However, both members of the couple can switch around after they sit down.
So the total arrangements where they sit together is
2! x 3 = 6
or
2 x P(3,2)
My answer would be
Total arrangements - arrangements where they sit together =
12 - 6 = 6
700+ your target then check out my 800 gmat blog here:
https://800gmatblog.zoxic.com
Top
User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Sun Nov 23, 2008 2:33 pm
rmpaes wrote:The easiest way to solve this problem is 6
It is a bold claim!
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Wed Sep 17, 2008 8:35 pm

by sukhpreet » Sun Nov 23, 2008 5:08 pm
Thanks logitech and rmpaes. Both answers made sense.
Top
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 am
Thanked: 27 times
Followed by:1 members
GMAT Score:660(

by 4meonly » Mon Nov 24, 2008 12:19 am
We have 4 seats
Let we have Mary and John as students
Mary cannot sit near John, so she must have at least 1 set near her
So we have Mary + 1 one seat. Let this pair will be X
Finally we have group of 3
X, J and one seat
they can be arranged in 3! ways - 6 ways

thats all
Top
Post new topic Post Reply
• Page 1 of 1