## kryton1212 one year ago 4 boys and 4 girls are arranged to sit in a row. Find the number of arrangements that can be made if : (a) there is no restriction, >>>(4+4)!=40320<<< (b) the boys must sit separately, >>>4!*4!=576<<< (c) 2 girls must sit at the ends. >>>(8-1)!*2!=10080<<< is it correct?

1. kryton1212

@ganeshie8

2. ganeshie8

a is correct b think again c wrong

3. ganeshie8

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!

4. kryton1212

oh i see...

5. kryton1212

how about part c then?

6. RolyPoly

Errr, how can you tie all boys in one bag if all boys must sit separately in part b?

7. kryton1212

@ganeshie8

8. RolyPoly

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.

9. RolyPoly

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'.

10. kryton1212

i understand your point. but i am still confusing part b and c...

11. RolyPoly

Which part(s) are you confused at?

12. kryton1212

b and c

13. RolyPoly

I mean what..

14. kryton1212

why tie in a bag is wrong?

15. RolyPoly

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

16. kryton1212

oh yes.

17. RolyPoly

Clear?

18. kryton1212

clear. so what should we do in this case?

19. RolyPoly

?

20. kryton1212

i mean, what should be the correct working step?

21. RolyPoly

22. kryton1212

i cannot understand 6 x.

23. RolyPoly

That's part c,right?

24. kryton1212

i got part b now. and go on to part c

25. RolyPoly

Those x's are either boys or girls.

26. RolyPoly

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.

27. digitalmonk

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

28. kryton1212

oh, i made it as (8-1)!*2! ... understand now...i forgot 2 girls must sit at the ends....

29. digitalmonk

:)