Does the last one mean that the alpha dog must be chosen and any other four animals?
If so, then .
There are 7 dogs and 13 cats. How many ways can 5 animals be selected if there must be exactly 2 dogs and 3 cats? What if there must be a least 2 dogs? When the alpha dog must be chosen?
Not really sure if my reasonings right, so if anyone could tell if my thinking is right, that'd be great.
When there must be 2 dogs and 3 cats:
7C2 = 21
13C3 = 286
Total # of combinations = 21 x 286
= 6006 combinations of cats and dogs
When there must at least be 2 dogs:
n(At Least 2 Dogs)' = n(1 Dog) + n(No Dogs)
= 13C4 + 13C5
= 715 + 1287
= 2002
n(Combinations of Any Type) = 20!/(7!x13!)
= 77520
n(At Least 2 Dogs) = n(Combinations of Any Type) - n(At Least 2 Dogs)'
= 77520 - 2002
= 75518
And for the last one, I have no clue how to start. Any help would be greatly appreciated!