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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. >>>(8-1)!*2!=10080<<< is it correct?

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  1. kryton1212
    • one year ago
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    @ganeshie8

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

  3. ganeshie8
    • one year 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
    • one year ago
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    oh i see...

  5. kryton1212
    • one year ago
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    how about part c then?

  6. RolyPoly
    • one year 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
    • one year ago
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    @ganeshie8

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

  11. RolyPoly
    • one year ago
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    Which part(s) are you confused at?

  12. kryton1212
    • one year ago
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    b and c

  13. RolyPoly
    • one year ago
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    I mean what..

  14. kryton1212
    • one year ago
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    why tie in a bag is wrong?

  15. RolyPoly
    • one year 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
    • one year ago
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    oh yes.

  17. RolyPoly
    • one year ago
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    Clear?

  18. kryton1212
    • one year ago
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    clear. so what should we do in this case?

  19. RolyPoly
    • one year ago
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    ?

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

  21. RolyPoly
    • one year ago
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    Read above comments

  22. kryton1212
    • one year ago
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    i cannot understand 6 x.

  23. RolyPoly
    • one year ago
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    That's part c,right?

  24. kryton1212
    • one year ago
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    i got part b now. and go on to part c

  25. RolyPoly
    • one year ago
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    Those x's are either boys or girls.

  26. RolyPoly
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    :)

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