A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

In how many arrangements can 3 boys and 4 girls stand in a row such that no two boys are together?

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ganeshie8 @ParthKohli @AriPotta

  2. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Number of ways that they can stand in - number of ways two boys can stand together. Note that the latter also covers the number of ways in which three boys can stand together.

  3. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, and also keep in mind that they can permute among themselves, like @ganeshie8 spotted the last time.

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got 5040 at total number of ways they can stand together

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How would you calculate the number of ways two boys can stand together? I feel like I am messing up there

  6. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Just like you did the last time.

  7. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, and yes, you have three boys, so you also need to see that you can choose any 2 out of 3 boys.

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    See I messed up somewhere I got 5028...

  9. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    4 girls can be arranged in 4! after that, 3 boys can be inserted in between the boys in 5C3 * 3! ways so in total we have 4! * 5C3*3! = 1440 ways

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    720 4,320 3,600 4,000

  11. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Why 5C3?

  12. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[7! - 2 \cdot \binom{3}{2}\cdot 6!\]

  13. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    place girls in the blanks first |dw:1432838542055:dw|

  14. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, I see.

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    after that, we have 5 places for placing boys |dw:1432838582145:dw|

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but 1440 is not an option. so what is wrong?

  17. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Is 720 correct? Just wondering...

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    720 is an option

  19. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Is the option correct?

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I dont think so?

  21. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What's the correct one?

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Im not sure :/

  23. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i don't see how it cannot be 1440

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1440 is that arrangement is the only the girls are sitting on the end seats

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Pattern: gbgbgbg # of arrangements: 4!*3! = 144

  26. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    girls need not be in the ends |dw:1432839047302:dw|

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    # of ways to fill the end seats: 4*3 = 12 # of ways to fill the other 5 seats: 5! # of arrangements: 12*5! = 1440

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah :/ I am getting the same way... but this is not an answer...

  29. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    as the question stands, 1440 is the correct answer it just so happens that the options don't have the correct answer... inform ur teacher and move on to next q

  30. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.