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lgbasallote

  • 3 years ago

How many ways can you arrange four people in a row of four seats? is this 4P4?

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  1. GOODMAN
    • 3 years ago
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    No, its 4!

  2. GOODMAN
    • 3 years ago
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    4!= 4*3*2*1

  3. sauravshakya
    • 3 years ago
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    There is only one way

  4. sauravshakya
    • 3 years ago
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    But If u are asking for different arrangement then its 4!

  5. lgbasallote
    • 3 years ago
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    \[4 P 4 = \frac{4!}{(4-4)!} \implies \frac {4!}{0!} \implies 4!\] @GOODMAN

  6. GOODMAN
    • 3 years ago
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    So why go through all the trouble?

  7. LolWolf
    • 3 years ago
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    Yep, \(_4P_4\) is correct.

  8. lgbasallote
    • 3 years ago
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    because of this next question im about to ask

  9. lgbasallote
    • 3 years ago
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    what about for arranging 4 people in 5 seats? is that 5P4? and arranging 5 people in 4 seats is also 5P4?

  10. sauravshakya
    • 3 years ago
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    Actually both will be 5!

  11. sauravshakya
    • 3 years ago
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    For first case it would be 5*4!=5! For second case it would be 5*4*3*2 = 5!

  12. sauravshakya
    • 3 years ago
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    If u are taking about arrangement

  13. lgbasallote
    • 3 years ago
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    ....so they are both 5P4?

  14. sauravshakya
    • 3 years ago
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    Yep

  15. sauravshakya
    • 3 years ago
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    I think so

  16. LolWolf
    • 3 years ago
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    Yes to both responses. To check we can show that: 4 people in 5 seats: We can always choose a single person whom is not in a seat, and there are 5 ways to do this, while the remaining 4 get a seat, thus: \(5\cdot_4P_4= _5P_4\), while for the latter, we have, simply \(_5P_4\) as the definition of "permute".

  17. lgbasallote
    • 3 years ago
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    i think im getting the hang of this

  18. LolWolf
    • 3 years ago
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    Sounds good

  19. karatechopper
    • 3 years ago
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    So...like 16?

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