Respuesta :
Answer:
They have 13800 different sequences of plays.
Step-by-step explanation:
The number of ways or permutations in which we can arrange x elements that are select from a group of n elements is calculated as:
[tex]nPx=\frac{n!}{(n-x)!}[/tex]
So, we have 25 different plays, and we need to choose 3 and make a sequence with them. Then, we can replace n by 25 and x by 3 and get:
[tex]nPx=\frac{25!}{(25-3)!}=13800[/tex]
Therefore, they have 13800 different sequences of plays.
Answer:
[tex] 25C3 = \frac{25!}{(25-3)! 3!}[/tex]
[tex] 25C3 = \frac{25!}{22! 3!} [/tex]
And using properties for the factorial term we can do this:
[tex] 25C3 =\frac{25*24*23*22!}{22! *3!}= \frac{25*24*23}{3*2*1}= 2300[/tex]
So then we will have 2300 possible sequences of plays with the conditions required
Step-by-step explanation:
For this case we can use the formula for cominatory since for this case theorder in which we select the 3 plays from the total of 25 is no matters. We can use the term (nCx) who means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And for this casr the different sequences of plays do they have are given by 25C3, replacing into the formula we got:
[tex] 25C3 = \frac{25!}{(25-3)! 3!}[/tex]
[tex] 25C3 = \frac{25!}{22! 3!} [/tex]
And using properties for the factorial term we can do this:
[tex] 25C3 =\frac{25*24*23*22!}{22! *3!}= \frac{25*24*23}{3*2*1}= 2300[/tex]
So then we will have 2300 possible sequences of plays with the conditions required
