Problem Solving

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.

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)

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

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

rmpaes wrote:The easiest way to solve this problem is 6

It is a bold claim!

Thanks logitech and rmpaes. Both answers made sense.

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