Probability (1 Viewer)

[1729]

How do you do part (iii) for this question? Thanks in advance

 

[kawaiipotato]

Count the ways to fix the boys in a circle first. Then count the ways to place the girls between the boys.
This will be your desired outcomes, and your total outcomes is just 7!.

 

[HeroWise]

But the question states "Chosen"

So if thats the case and we choose from the original 6 boys and 6 girls shouldnt you get this?

 

[InteGrand]

But the question states "Chosen"

So if thats the case and we choose from the original 6 boys and 6 girls shouldnt you get this?

The question's wording is such that it wants us to assume we have already picked 4 boys and 4 girls.

 

[HeroWise]

I see, i thought it was a carry over

 

[Roy G Biv]

i) 12C8
ii) 6C4 x 6C4
iii) 4!3! / 7!

Draw a circle with 8 spots.

We 'fix' in 1 boy. 3 boys left. Now, we're going to place the boys making sure to leave a seat empty between each boy. 3 boys can go in 1st of those seats, 2 in the second, 1 in the last.

Now we will fill in the blanks with the girls. 4 girls can go in first of the empty spots, 3 in the second, etc.
therefore: 4! x 3!

Since they've asked us the probability, we need to answer in terms of total possible ways to arrange these 8 students in a circle, which is 7!.

Hence 4!3! / 7!

Note: HeroWise has done this visually in his/her drawing above