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). 
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.
 A: 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 $$\frac 12 \cdot 7!=2520$$
A: Choose two of the seven positions for the cat and the rat.  Since the rat must be to the left of the cat, there is only one way to arrange the cat and the rat in these positions.  The five dogs can be arranged in the remaining five positions in $5!$ ways.  Hence, there are
$$\binom{7}{2}5!$$
ways to arrange the seven animals so that the rat is somewhere to the left of the cat.
A: 
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

Multiplying is correct only when the two choices are independent. You have six options for where to put the cat, and six options for where to put the rat, but you don't have both choices at the same time. If you put the cat all the way to the right, there are six ways to put the rat to the left of the cat. However, there is only one way to put the cat all the way to the right. So the total number of configurations where the cat is all the way to the right is $1*6$, not $6*6$. So you end up with $6+5+4+3+2+1 = 21$. When you multiply that by $5!$, you get $2520$, which matches the other answers.
A: Hint:
Suppose the rat is the leftmost animal. Than it is always left to the cat and you have $6!$ ways to permute the other animals.
Suppose the rat is in the 2nd place from the left. Again there are $6!$ permutations of the animals but if the cat is at the leftmost side your condition is not satisfied. So yuo have to subtract to $6!$ the number of ways in which the cat is at the first place. Theese are exactly $5!$ (you have 7 places, 2 fixed)
And so on. The total number is the sum of all the cases you have to analize.
