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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.
>>>(81)!*2!=10080<<<
is it correct?
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. >>>(81)!*2!=10080<<< is it correct?

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0a is correct b think again c wrong

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0for 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!

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0how about part c then?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0Errr, how can you tie all boys in one bag if all boys must sit separately in part b?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0When 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.

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0When 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'.

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0i understand your point. but i am still confusing part b and c...

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0Which part(s) are you confused at?

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0why tie in a bag is wrong?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0When 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

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0clear. so what should we do in this case?

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0i mean, what should be the correct working step?

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0i cannot understand 6 x.

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0i got part b now. and go on to part c

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0Those x's are either boys or girls.

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0You 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.

digitalmonk
 one year ago
Best ResponseYou've already chosen the best response.0when 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

kryton1212
 one year ago
Best ResponseYou've already chosen the best response.0oh, i made it as (81)!*2! ... understand now...i forgot 2 girls must sit at the ends....
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