anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
a is correct b think again c wrong
ganeshie8
  • 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!

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anonymous
  • anonymous
oh i see...
anonymous
  • anonymous
how about part c then?
anonymous
  • anonymous
Errr, how can you tie all boys in one bag if all boys must sit separately in part b?
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
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.
anonymous
  • anonymous
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'.
anonymous
  • anonymous
i understand your point. but i am still confusing part b and c...
anonymous
  • anonymous
Which part(s) are you confused at?
anonymous
  • anonymous
b and c
anonymous
  • anonymous
I mean what..
anonymous
  • anonymous
why tie in a bag is wrong?
anonymous
  • anonymous
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
anonymous
  • anonymous
oh yes.
anonymous
  • anonymous
Clear?
anonymous
  • anonymous
clear. so what should we do in this case?
anonymous
  • anonymous
?
anonymous
  • anonymous
i mean, what should be the correct working step?
anonymous
  • anonymous
Read above comments
anonymous
  • anonymous
i cannot understand 6 x.
anonymous
  • anonymous
That's part c,right?
anonymous
  • anonymous
i got part b now. and go on to part c
anonymous
  • anonymous
Those x's are either boys or girls.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
oh, i made it as (8-1)!*2! ... understand now...i forgot 2 girls must sit at the ends....
anonymous
  • anonymous
:)

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