Three boys and eight girls are seated randomly in a row of $11$ chairs, with all orders equally probable. what is the probability that there are at least $2$ girls between every pair of boys?

In the problem we are asked to find the probability that 2 girls are between a pair of boys. Hence we can generate the following arrangements:

b g g b g g b g g g g                           -- arrangement 1

In this situation, we are only interested on the part between the boys, hence we create a boundary just for the sake of calculation.

|b g g b g g b| g g g g                       -- arrangement 2

To solve this problem, we use the Combination formula: nCr = n! / n! (n-r)!

We can see in the above arrangement 2, that for the 3 boys there are 7 places which he can take, therefore:

7C3 = 7! / 7! (7-3)!

7C3 = 35

While the total places the 3 boys can take are 11 based on arrangement 1, therefore:

11C3 = 11! / 11! (11-3)!

11C3 = 165

Therefore the probability that two girls will be between two boys is:

Probability = 35 / 165

Probability = 0.2121

Therefore there is a 21.21% probability that 2 girls is between 2 boys.

50 min

step-by-step explanation:

they will be decreasing both their times in half

60 / 2 = 30

40/2 = 20

30 + 20 = 50

lol

step-by-step explanation: