Twelve coins are tossed and placed in a line. Each coin can show either a head or a tail.
Find the number of different arrangements of heads and tails which can be obtained.
I'm aware that that's a very easy question to solve. I just need to say 2! (The two possibilities for each coin) multiplied by it's self for 12 times (since there are twelve coins).
But I wanted to solve it in another way and I failed miserably.
What I did is, I imagined each head and tail to be a unique object
So there are H1 and T1 for first coin, there are H2 and T2 for second coin and so on. Eventually we end up with 24 objects.
Then I made these 24 objects change places in the 12 places available. (24p12)
Then since all heads are the same and all tails are the same, and there are 12 tails and 12 heads, I divided by 12!×12!, (24p12/ 12!×12!).
And you can already guess by now, my answer turned out waaaaaay smaller than the actual answer (5.6..×10^-3)
I thought about it and tried to imagine it with less number of coins and I think it turned like that because I assumed that each arrangement would contain the twelve heads and the twelve tails (apparently impossible) thus dividing by 12!×12!, when in reality an arrangement of 12 can contain either 3 heads and 9 tails so I would have to divide this unique arrangement by 3!×9! While another will have 2H and 3T…etc
So I guess the only way to be able to solve it in another way is to make a table where I'll write the number of possible tails and heads in an arrangement of 12 then multiple by 12! to make them change places then divide according to the number of tails and heads in that certain arrangement, which is time consuming.
Anyway I know there must be another way to solve this without saying (2!)^12 or doing my very wrong method, and I feel like this can be done using Combination but I don't know how.
Best Answer
First, the answer should be $2^{12}$, without factorial, although they are of the same value.
Second, saying $2^{12}$ directly is the cleanest way to go. There is, however, a way that forces the contribution of Combination.
Suppose there are $k$ heads (and therefore $12-k$ tails). How many different places can these $k$ heads be in?
By using combination/binomial coefficients, we should have $C^{12}_k$, or $\binom {12} k$.
There could be $0,1,\dots 12$ heads. Hence we add up all possibilities:
$$\binom{12}0+\binom{12}1+\cdots+\binom {12}{12} = \sum_{k=0}^{12}\binom{12}{k}$$
which is equal to $2^{12}$ as a consequence of the binomial theorem.
This is, incidentally, also a combinatorial proof of this identity: $\displaystyle \sum_{k=0}^n\binom nk = 2^n$.