Permutations with restrictions

Question

Five cows and one set of twin calves can be housed separately in a row of seven stalls in 7p7 ways. Find how many arrangements for:

(a) the two calves are not in adjacent stalls

(b) the two calves and their mother, who is one of the 5 cows, are in adjacent stall

Answer 1

I will work part a for you; you work part b in a similar manner.

If you put a calf in either end stall there 5 other stalls in which to put the other calf: 2 calves, 2 ends, 5 choices and then there are 5! arrangements of the cows.

If you put a calf in a stall of than an end there are 4 other stalls in which to put the other calf: 2 calves, 4 choices and then 5! arrangements for cows.

That means 20* 5! + 8*5! = 28*5!