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!.
How do you do part (iii) for this question? Thanks in advance
Screen Shot 2018-06-23 at 1.19.17 pm.png
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!.
Sxc avatar made by a sxc person: carrotontheground http://community.boredofstudies.org/...otontheground/
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?![]()
I see, i thought it was a carry over
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![]()
Last edited by Roy G Biv; 24 Jun 2018 at 12:02 PM.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks