In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together? Problem : 
In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row
so that no two lions are together?
1st Approach : 
L T L T  L T L T L
The 5 lions should be arranged in the 5 places marked 'L'
This can be done in 5! ways.
The 4 tigers should be in the 4 places marked 'T'
This can be done in 4! ways.
Therefore, the lions and the tigers can be arranged in 5! $\times$ 4! ways = 2880 ways.
2nd Approach : 
But I want to approach in the following way : 
Let all the lions appear together so that 4L = 1unit 
So, there are 4 tigers + 1 Lion = 5 units
5 things can be arranged in $5!$ ways and 5 lions can be arranged themselves in 5! ways , So there are 5! $\times 5!$ ways. 
Total number of ways in which we can arrange 9 items = 9! ways. 
Therefore condition when no two lions never appear together 
= 9! - $5! \times 5!$ = 5!( 9.8.7.6 - 5.4.3.2) = 348480 
But this wrong. Please suggest thanks...
 A: you have subtracted the cases where all the lions are together but you have to substract the case where 2 or 3 or 4 lions are together!!
A: The complement of "no two lions are together" is "at least two lions are together" since together, these two cases cover all possible arrangements (and don't overlap at all). This is very different to "all five lions are together".
A: First you are wrong, because you assume that all those are together, instead of looking as just a pairs of lion.
First find all possible combinations. Because all lions and tigers are distinct, then for the first place there are $(5+4)=9$ options, for the second $(5+4-1)=8$ options, and so on so the number of all posible combinations is:
$$9 \times 8 \times 7 \times \ldots \times 2 \times 1 = 9! = 362880$$
To get the exact solution you need to subtract the "bad" solution. So in order to find the bad solution, you could think of the 2 lions that are together as one. So there are 8 animals to rearrange. So there are $8! = 40320$.
Now we go back to the pair of lions. There are $5$ lions and there are $\binom{5}{2}$ ways to select a pair of lions so we have:
$$\binom 52 \times 8! = 403200$$
But that would give a negative amount, which is impossible. This happens because of double counting. Let the bold lions be pairs:
$L, T, L, $ L, L$, T, T, L, T$
$L, T, $ L, L$, L, T, T, L, T$
As you can in our cointing we count them as two, but they are actually one.
So in order to coun those "combinations" we would need to count special case, when we have 2 lions together, in between two tigers, then we have 3 lions together, in between two tigers. That's painful, but it get even more painful, because if those lions are at the end ot at the beggining they don't have to be in between two tigers. So I recommend you to stick to your first approach, because in combinatorics we can have multiple approaches, but we always should stick to the easiest one, that's your first approach.
You were lucky, because there is just one combination and that's:
$$L, T, L, T, L, T, L, T, L$$
If there were more you can add 2 "auxiliary" tigers, and then use the stars and bars calculation for $n-tuples$ with only positive integers. So we have $\binom{Tigers+2}{Lions}$ possible combinations. Then multiply by the number of permutations of the tigers and lions. So the final formula will look:
$$\binom{Tigers+2}{Lions} \times Tigers! \times Lions!$$
