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kryton1212

  • 2 years 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?

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  1. kryton1212
    • 2 years ago
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    @ganeshie8

  2. ganeshie8
    • 2 years ago
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    a is correct b think again c wrong

  3. ganeshie8
    • 2 years ago
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    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
    • 2 years ago
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    oh i see...

  5. kryton1212
    • 2 years ago
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    how about part c then?

  6. RolyPoly
    • 2 years ago
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    Errr, how can you tie all boys in one bag if all boys must sit separately in part b?

  7. kryton1212
    • 2 years ago
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    @ganeshie8

  8. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    i understand your point. but i am still confusing part b and c...

  11. RolyPoly
    • 2 years ago
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    Which part(s) are you confused at?

  12. kryton1212
    • 2 years ago
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    b and c

  13. RolyPoly
    • 2 years ago
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    I mean what..

  14. kryton1212
    • 2 years ago
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    why tie in a bag is wrong?

  15. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    oh yes.

  17. RolyPoly
    • 2 years ago
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    Clear?

  18. kryton1212
    • 2 years ago
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    clear. so what should we do in this case?

  19. RolyPoly
    • 2 years ago
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    ?

  20. kryton1212
    • 2 years ago
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    i mean, what should be the correct working step?

  21. RolyPoly
    • 2 years ago
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    Read above comments

  22. kryton1212
    • 2 years ago
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    i cannot understand 6 x.

  23. RolyPoly
    • 2 years ago
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    That's part c,right?

  24. kryton1212
    • 2 years ago
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    i got part b now. and go on to part c

  25. RolyPoly
    • 2 years ago
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    Those x's are either boys or girls.

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

  29. digitalmonk
    • 2 years ago
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    :)

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