There are 10 doors to a coliseum. In how many ways can a person go in and out if one can enter any door but leave using another door? Use permutation formula (P(n,r)) to solve.

Respuesta :

Answer:

90 is the answer

Step-by-step explanation:

P(10,2) = [tex]\frac{10!}{(10-2)!}[/tex] = [tex]\frac{10!}{8!}[/tex] = [tex]\frac{10x9x8!}{8!}[/tex]

10x9

=90

The number of ways should be 90.

Calculation of no of ways:

Since There are 10 doors to a coliseum.

So, here we used the permutation

[tex]= \frac{10!}{(10!-2!)}\\\\ = \frac{10!}{8!} \\\\= \frac{10\times 9\times 8!}{8!} \\\\= 10\times 9[/tex]

= 90

Learn more about permutation here: https://brainly.com/question/14767366?referrer=searchResults