Question

11. A football team has 25 different plays in their playbook and the coach

randomly chooses 3 different plays to start the game. How many different

sequences of plays do they have?

Answer 1

`25C3 = (25!)/((25-3)! 3!)`

`25C3 = (25!)/(22! 3!)`

And using properties for the factorial term we can do this:

`25C3 =(25*24*23*22!)/(22! *3!)= (25*24*23)/(3*2*1)= 2300`

So then we will have 2300 possible sequences of plays with the conditions required

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:

`nCx=(n!)/((n-x)! x!)`

And for this casr the different sequences of plays do they have are given by 25C3, replacing into the formula we got:

`25C3 = (25!)/((25-3)! 3!)`

`25C3 = (25!)/(22! 3!)`

And using properties for the factorial term we can do this:

`25C3 =(25*24*23*22!)/(22! *3!)= (25*24*23)/(3*2*1)= 2300`

So then we will have 2300 possible sequences of plays with the conditions required

Answer 2

They have 13800 different sequences of plays.

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:

`nPx=(n!)/((n-x)!)`

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:

`nPx=(25!)/((25-3)!)=13800`

Therefore, they have 13800 different sequences of plays.