Answer :
We have to use basic combinatorics to solver his problem. We first have to determine what kind of problem this is.
Does order matter in this question? Yes it does. So it is a permutation.
Is is a permutation with repetition? No, because once a performer goes, they cannot perform again. Hence, we have defined this problem as a permutation without repetition. The formula for this kind of question is:
[tex] \frac{n!}{(n-r)!} [/tex]
Where n is the number of things we choose from, and p is the number of places we want to put them in.
There are 4 places for 4 performers, because 1st performer, lets call him Joe, has already taken the 1st spot. Hence, we substitute our values:
[tex] \frac{4!}{(4-4)!} [/tex]
= [tex]4![/tex]
=[tex]24[/tex]
There are 24 different ways to schedule this performance.
Hope this helped!
Sincerely,
~Cam943, Junior Moderator
Does order matter in this question? Yes it does. So it is a permutation.
Is is a permutation with repetition? No, because once a performer goes, they cannot perform again. Hence, we have defined this problem as a permutation without repetition. The formula for this kind of question is:
[tex] \frac{n!}{(n-r)!} [/tex]
Where n is the number of things we choose from, and p is the number of places we want to put them in.
There are 4 places for 4 performers, because 1st performer, lets call him Joe, has already taken the 1st spot. Hence, we substitute our values:
[tex] \frac{4!}{(4-4)!} [/tex]
= [tex]4![/tex]
=[tex]24[/tex]
There are 24 different ways to schedule this performance.
Hope this helped!
Sincerely,
~Cam943, Junior Moderator
