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There are 6 dogs and 5 cats.

In how many different orders can these animals be placed in line if any animal can be next to any other animal?

In how many different orders can these animals be placed in line if the dogs and cats are lined up alternately?
(Hint - The first animal MUST be a dog)

In how many different orders can these animals be placed in line if the first and last animal in line must be a cat?

Sagot :

Using the arrangements formula, the number of orders is given as follows:

  • 39,916,800 if no restrictions.
  • 86,400 if they are lined up alternatively.
  • 7,257,600 if the first and last must be cats.

What is the arrangements formula?

The number of possible arrangements of n elements is given by the factorial of n, that is:

[tex]A_n = n![/tex]

When there are no restrictions, the number of ways is:

[tex]A_{11} = 11! = 39,916,800[/tex]

When they must be lined alternatively, the 6 dogs can be arranged in 6! ways, and the 5 cats in 5! ways, hence the number of orders is:

[tex]A_6A_5 = 6! \times 5! = 86,400[/tex]

When the first and last are cats, we have that:

  • For the first and last animals, there are 5!/2! = 20 ways.
  • For the middle 9 animals, there are 9! ways.

Hence:

20 x 9! = 7,257,600.

More can be learned about the arrangements formula at https://brainly.com/question/24648661

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