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…
Best Answer
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!!