There are 11 portable mini suites (a. k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with their guests' names. There are 5 poodles and 6 tabbies.

How many ways can the "suites" be arranged if:

a. there are no restrictions.

b. cats and dogs must alternate.

c. dogs must be next to each other.

d. dogs must be next to each other and cats must be next to each other.

Step-by-step explanation:

a) 7!

If there are no restrictions, answer is 7! as it is the permutation of all animals.

b) 4! x 3!

As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among  themselves and the order of the dogs among themselves. There are 4! permutations of the cats and 3! permutations of the dogs, so there are a total of 4! x 3! possible arrangements of the suites.

c) 3! x 5!

There are 3! possible arrangements of  the dogs among themselves. Now, if we consider the dogs as  one ”object” together, then we can think of arranging the 4 cats  together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.

d) 2 x 4! x 3!

As required that all the cats must be together and all the dogs must be together, either the cats are all  before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e.  2 x 4! x 3!

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

answer: c. is your answer

step-by-step explanation: get me a black leather coat