lgbasallote
How many ways can you arrange four people in a row of four seats?
is this 4P4?
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GOODMAN
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No, its 4!
GOODMAN
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4!= 4*3*2*1
sauravshakya
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There is only one way
sauravshakya
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But If u are asking for different arrangement then its 4!
lgbasallote
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\[4 P 4 = \frac{4!}{(4-4)!} \implies \frac {4!}{0!} \implies 4!\]
@GOODMAN
GOODMAN
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So why go through all the trouble?
LolWolf
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Yep, \(_4P_4\) is correct.
lgbasallote
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because of this next question im about to ask
lgbasallote
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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?
sauravshakya
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Actually both will be 5!
sauravshakya
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For first case it would be 5*4!=5!
For second case it would be 5*4*3*2 = 5!
sauravshakya
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If u are taking about arrangement
lgbasallote
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....so they are both 5P4?
sauravshakya
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Yep
sauravshakya
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I think so
LolWolf
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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".
lgbasallote
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i think im getting the hang of this
LolWolf
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Sounds good
karatechopper
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So...like 16?