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?
Best Answer
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 $5040-2\cdot720 = 3600$.