A probability problem by Arpan Sarangi

Probability Level 2

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. What is the value of $\frac{m}{n}$ ?

6 7 8 4 5

2 solutions

Halum Singh
Aug 12, 2015

Taking 5 girls as a unit, we can arrange the girls with the boys in 6 ! ways. In each of the cases girls in the unit can be arranged in 5 ! ways.
So, n = (6 !).(5 !)
In second case, we can choose four girls in 5 ways and for each choice they can stand in the queue in 6 X 4 ! ways(4 position is in between boys + extreme two side position ) and for each case 5th girl can be stand in 5 ways and boys can arrange among themselves in 5 ! ways.
So, m = 5.6.(4 !).5.(5 !) = 5. (6 !). (5 !) So, m/n = 5.

Louis W
Aug 8, 2015

I am unfamiliar with the formula necessary to answer this, but it took a quarter of a piece of paper to just do this.

First with 5 girls, there can be:

5 behind

1 ahead and 4 behind

2 ahead and 3 behind

3 ahead and 2 behind

4 ahead and 1 behind

5 ahead

For a total of n=6.

Then with 4 girls, the trick here is that EXACTLY 4 girls are consecutive, so one of the options for the fifth girls is not next to the other 4 girls. There can be:

6 behind, the girl in any of the 5 latter spots (5)

1 ahead and 5 behind, the girl in any of the latter 4 spots (4)

2 ahead and 4 behind, the girl in front or any of the latter 3 spots (4)

3 ahead and 3 behind, the girl in any of the front or latter 2 spots (4)

4 ahead and 2 behind, the girl in any of the front 3 or last spot (4)

5 ahead and 1 behind, the girl in any of the front 4 spots (4)

6 ahead, the girl in any of the front 5 spots (5)

The number in parenthesis has been the total possibilities for that option, so m=30.

$\frac {m}{n}=5\space\space\space\Box$

Had the numbers in the problem been greater, like 20 boys and 20 girls, some formula would have to come into play to not waste a bunch of paper.

Can you explain your 2nd case clearer? I don't see why there are only 4 possibilities after that. I'm also not too clear on what you mean by "6 behind, the girl in any of the 5 latter spots".

Calvin Lin Staff - 5 years, 10 months ago

The second part of the questions states that there are 5 boys and 5 girls standing in a line. There are exactly 4 girls standing consecutively. So much like the first part of my answer, these 4 consecutive girls can be standing in various places in line, and the other 6 people will be positioned accordingly.

In the first line, with 4 consecutive girls there can be 6 people behind them. This would imply that the 4 consecutive girls are in the front of the line and everyone else in behind them. In the first case all these other people were boys so it didn't matter their order. But in this case one of those people in a girl, so the girl's position matters. She cannot be next to the girls or there wouldn't be exactly 4 consecutive girls, so she can be in any of the 5 latter spots, for a total of 5 possibilities for that option.

The remaining options follow this idea.

Louis W - 5 years, 10 months ago

A probability problem by Arpan Sarangi

2 solutions

0 pending reports