How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).
I started out by arranging the 5 different dogs which is $5!$, and from here basically I got stuck, I tried to do it by cases, like if I arranged the 5 dogs, I have 6 places to put the cat in,starting from the right, the rat also can be put in 6 places (can be near the cat from the left), and moving one to the left each case, so I got: $6*6+5*5+4*4+3*3+2*2+1*1=91$ and this multiplied by the number of ways to arrange the dogs gave me the answer: $91*5!=10920$, now this seems clearly wrong and I messed up since $7!=5040$, which is the number of ways to arrange all seven, I am trying to understand my mistakes and how to deal with this question.
Thanks in advance to any help.
Best Answer
As you say there are $7!=5040$ arrangements if we ignore the restriction on the cat and rat. We can match up arrangements with the rat left of the cat with arrangements with the rat right of the cat, so the arrangements with the rat left of the cat are half the total. The result is $$\frac 12 \cdot 7!=2520$$