Among them are Miss Rabbit, Granddad Dog, Candy Cat, Peppa Pig, Grandpa Pig, Granny Pig, George Pig, Mummy Pig, and Daddy Pig. A resident who is infected from being outside can only infect one healthy person, so (a) how many ways can all of the residents get infected?
(b) How many ways are there for everyone to become infected if Miss Rabbit and Candy Cat do not go outside?
Imagine the following scenario: there are six characters seated at a table, three on each side, what is the probability that Grandpa Pig and Granny Pig will sit together?
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Answer
Let there be nine places such that one of them is infected at every iteration. The one on the first position gets infected first, then the one on the second position, and so on.
There are nine animals living together.
The following represents the order of infection. First place means Miss Rabit was infected first.
Number of ways they all get infected = total arrangements of 9 animals at the 9 spots = 9! = 362,880
The remaining 7 animals can take the 1st place if Miss Rabbit and Candy Cat don’t step outside.
Total ways = 7*8*7*6*5….*1 = 7*8! = 282,240
Grandpa pig and Granny pig sit together out of six people.
We first choose one side and then any one pair out of two pairs (12, 23 only, as 1,3 are not together) for them: 
= 4
Then we arrange remaining 4 people on the 4 seats along with the interarrangement of Grandpa pig and Granny pig : 4!*2! = 48
Total ways = 48*4 = 192
Total arrangement of 6 animals on 6 places = 6! = 720
P(Grandpa pig and Granny pig together) = 192/720 = 0.267
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