Today I was working on a problem, for which I think there are two possible answers to it,
the question is we need to Arrange five $0$'s and five $1$'s such that no two $0$'s come together and no two 1's come together. According to my understanding I am confused between two approach of getting answer,
(1)First Approach:
a)Firstly I will arrange five $1$'s in 5! ways as all $1$a's are same, not unique so i will divide it by 5!, so I get result as $1$ and similarly then arrange five $0$'s in 5! ways/5! i will get 1, so finally 1 in case my final number is starting with 1 and if it is starting with $0$ then I will have another $1$, so final answer is $1+1 = 2.$
(2)Another way I look at this problem, as first i will arrange five $1$'s in 5! ways/5! and then from spaces between these 1 which are 6, I will choose 5 space for 5 $0$'s to arrange, like this 6C5 and then arrange five $0$'s in $\frac{5!}{5!}$ ways. Similarly in case when final number starts with $0$.
so finally answer here I will get $6+6=12.$
So, my question where I am wrong in my second approach?, I know correct answer is $2$, but this second approach fails to answer this?
There are two problems with this approach. The first is that whatever the enumeration is for the $1$'s, once these are placed, then the positions of the $0$'s would be locked.
So, if the first portion of your method-2 answer was valid, which it is not, then the enumeration would be $6$, rather than $12$.
Now, examine the first portion of your method-2 answer.
--- 1 --- 1 --- 1 --- 1 --- 1 ---
The above diagram represents placing the $5$ ones, and indicates that you have $6$ available spaces before, after, and between the $1$'s. At the start of your method-2, you are assuming that you can select any one of these spaces, discard it, and place the $0$'s in the $5$ remaining spaces.
This is not true, because you are not allowed to have two consecutive $1$'s. Therefore, of the $6$ available spaces, the $4$ middle spaces must not be chosen. That is, if you choose one of the middle spaces as a place to not put a $0$, then you end up with two consecutive $1$'s, in the middle of the sequence.
So, with this correction, you can only choose one of the two outer spaces as the one that will not receive a zero. Then, as discussed, the positions of both the $1$'s and $0$'s are then locked.
Therefore, with these corrections in mind, method-2 also leads to the enumeration of $(2)$.