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No, its 4!

4!= 4*3*2*1

There is only one way

But If u are asking for different arrangement then its 4!

\[4 P 4 = \frac{4!}{(4-4)!} \implies \frac {4!}{0!} \implies 4!\]
@GOODMAN

So why go through all the trouble?

Yep, \(_4P_4\) is correct.

because of this next question im about to ask

Actually both will be 5!

For first case it would be 5*4!=5!
For second case it would be 5*4*3*2 = 5!

If u are taking about arrangement

....so they are both 5P4?

Yep

I think so

i think im getting the hang of this

Sounds good

So...like 16?