First the denominator. There are 8*7*6*5*4*3*2*1 = 40320 ways the 8 friends could be seated.

Now the numerator: There a 6 possible seats for Trevor (he cant have an end seat). For each of these 6 seats, the two brothers could be seated in 2 ways. (A on the left B on the right or A on the right and B on the left).
So 12 ways in which the 3 brothers could be seated. Finally, the remaining 5 seats could be filled 1*2*3*4*5=120 ways. So the numerator is 2*6*120=1440.

Please help me on this question:

Trevor and his 2 brothers and 5 friends are seated at random in a row of 8 seats at the cinema. What is the probability that Trevor has one brother on his immediate left and one on his immediate right?

I think it's hard! Thanks

Oh my goodness! I really did need help on this. Thank you very much appreciated =).

1/28 this is the answer I've come across this question before and the answer before

in how many ways can 10 people fill 3 chairs at a table

For the Trevor and his 2 brothers and 5 friends question:

The idea is Trevor cannot sit at the end seats ie seat no. 1 and 8. Imagine sitting at seat 1, you can only have people sitting to your right. (Assuming the seats are arranged like this: s1, s2, s3, s4, s5, s6, s7, s8).

Suppose Trevor sits on seat no. 2. Then of the 8 seats there are only 7 available since Trevor is occupying one by sitting on it. Therefore the first brother will have 1/7 chance of sitting next to him on one side. Now 6 chairs remain. So the other brother will only have 1/6 chance of sitting on the opposite side. Therefore overall: (1/8)*(1/7)*(1/6) = 1/336
Remember: Trevor sat on seat no. 2 but there are 6 seats which he could have sat on (2,3,4,5,6,7) and it would have still worked. So 6*(1/336)= 6/336.
Now one last thing: suppose
Brother 1 = B1
Brother 2= B2
Trevor = T
They can sit like so:
B1, T, B2
This is what we have worked out so far in the above working. The last thing is that they could have sat in the other way i.e.
B2, T, B1
So in total we have to consider 2 different ways of sitting as both would work.
Therefore, 2*(6/336) = 12/336
Simplifying 12/336 = 4/112= 2/56= 1/28

To solve this problem, we need to find the probability of Trevor having one brother on his immediate left and one on his immediate right.

Let's break down the problem and calculate the probability step by step:

Step 1: Calculate the total number of possible seating arrangements for the 8 friends.
We can use the formula for permutations, which is the product of the number of choices at each step.

In this case, there are 8 seats, so the total number of possible arrangements is 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320.

Step 2: Calculate the number of favorable seating arrangements.
To have one brother on Trevor's left and one on Trevor's right, we can break down the arrangements into three parts:
- Trevor's seat: Trevor has 6 possible seats (any seat except the two end seats).
- The two brothers' seats: Each brother can sit either on the left or right side of Trevor. So for each of Trevor's seat choices, there are 2 possible arrangements for the brothers.
- The remaining 5 seats: The 5 friends can sit in any of these 5 seats, resulting in 5 x 4 x 3 x 2 x 1 = 120 arrangements.

Therefore, the number of favorable seating arrangements is 6 x 2 x 120 = 1440.

Step 3: Calculate the probability.
The probability is the ratio of the number of favorable arrangements to the total number of possible arrangements.

Probability = Number of favorable arrangements / Total number of possible arrangements

Probability = 1440 / 40320

Simplifying this fraction, we get:

Probability = 1 / 28

So the probability that Trevor has one brother on his immediate left and one on his immediate right is 1/28.

Therefore, the answer to the question is 1/28.