1. ## Probability: Arrangements Question

Question: 5 people designated as A, B, C, D, E are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that

a) there is exactly one person between A and B

b) there is exactly 2 people between A and B

c) there is exactly 3 people between A and B

I am not sure how to work out this problem. I am having trouble coming out with the possible ways of outcomes and I do not know what to do with them. Here is what I have so far:

a) there is exactly one person between A and B

There is 3 ways that will happen ACBDE, ADBCE, AEBCD
but will ACBED count as another way?

Then there is 120 ways that these 5 people can line up.

Would we have to divide 3 by 120 to get the proability?

Thank you very much for your help.

2. Originally Posted by hotmail590
Question: 5 people designated as A, B, C, D, E are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that

a) there is exactly one person between A and B

b) there is exactly 2 people between A and B

c) there is exactly 3 people between A and B
(a)The possibilities are,
AXBXX This is (1)(3)(1)(2)(1)=6
BXAXX This is (1)(3)(1)(2)(1)=6
XAXBX ................................
XBXAX
XXAXB
XXBXA
Total=6 (times total arragement =6)=36

The number of ways to arrange 5 people is
(5)(4)(3)(2)(1)=120
Thus, probability is 36/120=3/10
---------
(b)
The possibilities are,
AXXBX This is (1)(3)(2)(1)(1)=6
BXXAX This is (1)(3)(2)(1)(1)=6
XAXXB ....................................
XBXXA
Total=6(times total arragement=4)=24
The number of ways to arrange 5 people is
(5)(4)(3)(2)(1)=120
Thus, the probability is 24/120=1/5=2/10
---------
(c)This only happens when,
AXXXB =6
BXXXA =6
In total we have 12, thus the probability is,
12/120=1/10