There are 7 acts in a talent show.
An acrobat, a comedian, a dancer, a guitarist, a juggler, a singer, and a violinist.
A talent show host randomly schedules the 7 acts.
Compute the probability of each of the following events.

Event A: The first four acts are the singer, the juggler, the guitarist, and the violinist, in any order.

Event B: The comedian is first, the guitarist is second, the dancer is third, and the juggler is fourth.

Write your answers as fractions in simplest form.

The number of different four acts taken from a set of seven acts, when the order does not matter, is calculated using the concept of combinations.

Thus, the number of ways that the first four acts can be scheduled is:

[tex]C(m,n)=\dfrac{m!}{n!(m-n)!}[/tex]

[tex]C(7,4)=\dfrac{7!}{4!(7-4)!}=\dfrac{7!}{4!(3)!}=35[/tex]

And the number of ways that four acts is the singer, the juggler, the guitarist, and the violinist, in any order, is 1: C(4,4).

Therefore the probability of Event A is:

[tex]P(A)=1/35[/tex]

Event B

Now the order matters. The difference between combinations and permutations is ordering. When the order matters you need to use permutations.

The number of ways in which four acts can be scheculed when the order matters is:

[tex]P(m,n)=\dfrac{m!}{(m-n)!}[/tex]

[tex]P(m,n)=\dfrac{7!}{(7-4)!}=P(m,n)=\dfrac{7!}{4!}=840[/tex]

The number of ways the comedian is first, the guitarist is second, the dancer is third, and the juggler is fourth is 1: P(4,4)

Therefore, the probability of Event B is:

[tex]P(B)=1/840[/tex]