Zoo 4 lions 7 tigers animal Counting problem, cage arrangement with restriction There are 4 lions and 7 tigers in a zoo. Each of them is put into a cage and the 11 cages are arranged in a row. Find the number of possible arrangements if only 3 lions are arranged next to each other, and the animals in the leftmost cage and the rightmost cage are tigers.
Attempt:
Each animals are assumed to be different.
The left term is that,
firstly 2 tigers are picked  from the 7 tigers.
Then group 3 lions together, picked from the 4.
Within the group of lion, they can arrange themselves so there would be $P_3^4$
So in the middle there would be 5 tigers, 1 lion and a group of 3 lions, total 7 entries.
In order to eliminate the case where 4 lions are together, the right term are subtracted from the left.
firstly 2 tigers are picked from the 7 tigers.
then all 4 lions are grouped together.
So in the middle there would be 5 tigers and a group of 4 lions, total 6 entries.
$$P_2^7 * P_7^7 * P_3^4 - P_2^7 * P_6^6 * P_4^4$$
$$= 42 * 5040 * 24 - 42 * 720 * 24$$
$$= 4354560.$$
But the correct answer is $3628800.$
I am unable to figure out why, could anyone help please?
 A: The error is that there are twice as many ways to get a 4-lions-in-a-row situation as are being counted here.
To see this, let's consider a situation in which all four lions are together, in the order $ABCD$.  There's two ways to get this in the group-of-three-lions method: we could have had $ABC$ as the group of three lions, and $D$ alone, and $D$ is to the right of $ABC$... or we could have $BCD$ as the group of three lions, and $A$ alone, and $A$ is to the left of $BCD$.  Both of these scenarios are counted once in the three-lions method, so both must be removed when excluding this case: you have $5040 - 720 = 4320$ cases (with indistinguishable creatures) but it should be $5040-2\cdot720 = 3600$.
A: The problem with the first term of your answer is that it double counts cases where $4$ lions are together. When you choose $3$ lions, arrange them and then add the $4th$ lion, it leads to duplicate arrangements. Take an example,
Say, we first choose lions $L1 ~ L2 ~ L3$ as one block and then later place $4th$ lion $L4$, we have an arrangement which has all lions together and in order $L1 ~ L2 ~ L3 ~ L4$.
Now if we chose $L2 ~ L3 ~ L4$ as a block and then later placed left out lion $L1$, we again would have an arrangement where all lions are together and in order $L1 ~ L2 ~ L3 ~ L4$.
Can you see the double counting?
But when you subtract, you do not subtract duplicate cases. So if you subtract $ 2 \cdot 42 \cdot 720 \cdot 24$ instead, you do get the right answer.

Now here is another approach - first take $3$ lions together and $7$ tigers, permute them in $8!$ ways. But these will have arrangements where lions are at two ends. So we subtract those arrangements. That gives us $(8! - 2 \cdot 7!)$ arrangements.
As the last lion cannot be at either end and not next to other $3$ lions, that leaves $5$ places for the last lion. Finally lions can be arranged internally in $4!$ ways.
That leads to total count of permissible arrangements as,
$(8! - 2 \cdot 7!) \cdot 5! = 3628800$
A: The stated answer is the number of ways of arranging seven tigers and four lions with tigers at each end of the row so that exactly three lions are together.  To see this, first arrange the seven tigers in a row, which can be done in $7!$ ways.  Since there are tigers at each end of the row, this leaves six spaces between successive tigers in which the lions can be placed.
$$t_1 \square t_2 \square t_3 \square t_4 \square t_5 \square t_6 \square t_7$$
Choose one of these six spaces to receive three lions and one of the five remaining spaces to receive the other lion.  Arrange the four lions from left to right in the selected spaces, which can be done in $4!$ ways.  Thus, there are
$$7! \cdot 6 \cdot 5 \cdot 4! = 7!6! = 10! = 3,628,800$$
ways to arrange the tigers and lions with the restrictions stated above.
