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kryton1212
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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?
 8 months ago
 8 months ago
kryton1212 Group Title
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?
 8 months ago
 8 months ago

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kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
@ganeshie8
 8 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
a is correct b think again c wrong
 8 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
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!
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
oh i see...
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
how about part c then?
 8 months ago

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

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
@ganeshie8
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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.
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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'.
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i understand your point. but i am still confusing part b and c...
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Which part(s) are you confused at?
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
b and c
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I mean what..
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
why tie in a bag is wrong?
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
oh yes.
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
clear. so what should we do in this case?
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i mean, what should be the correct working step?
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Read above comments
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i cannot understand 6 x.
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
That's part c,right?
 8 months ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i got part b now. and go on to part c
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Those x's are either boys or girls.
 8 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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.
 8 months ago

digitalmonk Group TitleBest ResponseYou've already chosen the best response.0
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
 8 months ago

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