Gaj81 wrote:5 boys and 4 girls are going for a movie. Andy and Sally are part of this group. For a restriction that no two boys or girls can sit beside each other, what is the probability of Andy sitting besides Sally?

Since no boy can sit next to another boy, the arrangement looks like this:
B-G-B-G-B-G-B-G-B

Case 1: Andy sits on EITHER END
P(Andy sits on either end) = 2/5. (Of the 5 boys' seats, 2 are on the end.)
P(Sally sits next to Andy) = 1/4. (Once Andy has been placed on either end, Sally must select the seat next to Andy, giving her only 1 good option from the 4 girls' seats.)
Since we want both events to happen, we MULTIPLY the fractions:
2/5 * 1/4 = 2/20 = 1/10.

Case 2: Andy sits in one of the MIDDLE SEATS
P(Andy sits in the middle) = 3/5. (Of the 5 boys' seats, 3 are in the middle.)
P(Sally sits next to Andy) = 2/4. (Once Andy has been placed, Sally must sit directly to the left or right of Andy, giving her 2 good options from the 4 girls' seats.)
Since we want both events to happen, we MULTIPLY the fractions:
3/5 * 2/4 = 6/20 = 3/10.

Since either Case 1 or Case 2 will yield a favorable outcome, we ADD the probabilities:
1/10 + 3/10 = 4/10 = [spoiler]2/5[/spoiler].

Gaj81 wrote:5 boys and 4 girls are going for a movie. Andy and Sally are part of this group. For a restriction that no two boys or girls can sit beside each other, what is the probability of Andy sitting besides Sally?

OA-2/5

Many probability questions can be solved intuitively rather than with long calculations. Let's try to do so with this problem!

We can get off to a great start by flipping around the question: what's the probability that Sally sits beside Andy?

We know that it has to go BGBGBGBGB. We can see that each girl sits beside 2 of the boys. Since there are 5 boys, there's therefore a 2/5 chance that Sally sits beside Andy - done!

Just from curiosity, how would you solve this one using permutations calculation?

Milovan wrote:Just from curiosity, how would you solve this one using permutations calculation?

Since the 5 boys must be separated, the arrangement looks like this:
B-G-B-G-B-G-B-G-B.

Total possible arrangements:
Number of ways to arrange the 5 boys = 5!.
Number of ways to arrange the 4 girls = 4!.
To combine these options, we multiply:
5!*4!.

Good arrangements:
In a good arrangement, Andy is next to Sally.
Number of options for Sally = 4. (Any of the 4 even-numbered seats.)
Number of options for Andy = 2. (To the left or right of Sally.)
Number of ways to arrange the remaining 3 girls = 3!.
Number of ways to arrange the remaining 4 boys = 4!.
To combine these options, we multiply:
4*2*3!*4!

P(Andy is next to Sally) = (good arrangements)/(total possible arrangements) = (4*2*3!*4!)/(5!*4!) = 2/5.