Movie time!

Probability Level pending

Four girls and four boys all sit in a row (of 8 seats) at a movie theater.

If they all sit down randomly, the probability that at least one girl will be seated next to two other girls (like the third girl in the picture) is a b \frac{a}{b} , where a a and b b are coprime positive integers. What is a + b a+b ?

Image credit: https://www.pinterest.com


The answer is 19.

Geoff Pilling by Geoff Pilling , 53, USA

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2 solutions

Geoff Pilling
Feb 6, 2017

If we don't distinguish between the boys or girls, there are 25 25 ways that they can sit such that one girl will be sitting next to two other girls.

This is calculated as follows... Suppose you had one girl, one "mega-girl" ( 3 3 girls together) and four boys. Then, the total number of ways they could sit would be given by 6 ! 4 ! = 30 \frac{6!}{4!} = 30 . Now, you have double counted 5 5 times, one for each of the ways that four sit together. So the total number of ways is 30 − 5 = 25 30 - 5 = 25 .

And the total number of ways they can sit is ( 8 4 ) = 70 \binom{8}{4} = 70 .

So, the probability is 25 70 = 5 14 \frac{25}{70} =\frac{5}{14}

5 + 14 = 19 5+14 = \boxed{19}

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Newly coined word: "mega-girl".... That did make me chuckle. :D

Anyway, nice approach. I took the complementary approach and looked at the 3 cases: (i) all girls separate, (ii) one pair of girls and two separate, and (iii) two pairs of girls. This resulted in 5 + 3 ∗ 10 + 10 = 45 5 + 3*10 + 10 = 45 arrangements where there isn't a "mega-girl", and thus 70 − 45 = 25 70 - 45 = 25 where there is one.

Brian Charlesworth - 4 years, 4 months ago

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Haha... Yup, "mega-girl"... Couldn't think of a more descriptive word at the time...

Myself, I've never had the fortune of dating one, but I could imagine they could be a bit over-bearing at times! ;0)

Geoff Pilling - 4 years, 4 months ago
Aman Dubey
Feb 14, 2017

This can happen in two ways

  1. We fill the "Mega girl" and another girl in between 4 boys so that the Mega girl and other girl dont sit next to each other and this could be done in x= 4!×(5C2)×2×4!

  2. We take all four girls together and arrange them in y=5!×4! ways. So probablity is (x+y)÷8!= 5÷14

5+14=19

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