At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
a is correct b think again c wrong
for b :- (b) the boys must sit separately, tie all 4 boys in one bag remaining 4 girls separate. so total 1 bag + 4 girls = 5 objects can be permuted in 5! ways boys inside bag can be permuted in 4! ways so total arrangements = 5!*4!
oh i see...
how about part c then?
Errr, how can you tie all boys in one bag if all boys must sit separately in part b?
When boy must sit separately, girls must also sit separately, e.g.: B G B G B G B G Then 4!4! is correct for this case. That's one arrangement. But you can also have a girl taking the first seat, i.e. G B G B G B G B So, you need to multiply your original answer by two.
When two girls must sit at two ends, you have: \(G_1, x, x, x, x, x, x, G_2\) So, you need to permute the 6 x's and the two G'.
i understand your point. but i am still confusing part b and c...
Which part(s) are you confused at?
b and c
I mean what..
why tie in a bag is wrong?
When you tie them together, they are connected and are no longer seperated, i.e. You take A = B B B B, and you arrange A G G G G
clear. so what should we do in this case?
i mean, what should be the correct working step?
Read above comments
i cannot understand 6 x.
That's part c,right?
i got part b now. and go on to part c
Those x's are either boys or girls.
You only need to fix two girls (of course you need to permute that two girls as well) at the end, the others, you can mix and match, let the rest permute themselves.
when two girls sit in the two ends they can do it in 2 ways so remaining 4 boys and 2 girls = 6 students can sit in 6! = 720 ways total no of ways = 2x720 = 1440
oh, i made it as (8-1)!*2! ... understand now...i forgot 2 girls must sit at the ends....