In how many ways can 10 instructors be assigned to ten sections of a course in mathematics?

Question

anonymous · Answer

Is it 1???

anonymous · Answer

@lgbasallote

anonymous · Answer

It's not a combination, it's a permutation.  If you were just choosing instructors to teach, then it would be combinations, but since you're ordering them into sections, it's permutations.

anonymous · Answer

idk not so good with this

lgbasallote · Answer

oh well then it's $$\frac{n!}{(n-r)!}$$?

anonymous · Answer

Yes,$$\huge ^{10}P_{10}$$

It would be:

anonymous · Answer

I solve using this method.  Put a blank for every section.

__ / __ / __ / __ / __ / __ / __ / __ / __ / __

So how many different instructors could teach the first class?  Well, there are 10, so it could be any one of them.  I put a 10 in the blank.  Then how many instructors could teach the second class?  Well, I've already chosen an instructor, so there are only 9 left.  I continue doing this for all sections and then multiply all the numbers.

anonymous · Answer

And the formula is:

$$\huge \color{green}{ ^{n}P_n = n!}$$

lgbasallote · Answer

it sounded like a selection to me though

anonymous · Answer

first section can be assigned one of the 10 teachers....2nd section one of the 9....3 section 8..............and finally 1st section only 1...

which is equal to 10*9*8*.......*2*1=10!

anonymous · Answer

so I count down then multiply

anonymous · Answer

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