# Why do we not use combinations here?

Ok I was going through the question:

Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is?

This might be elementary but Idk I am just stuck here... I checked the solution of the answer which was given in the following way:

Let's arrange six identical coins in a row. \
For the number of heads and tails to be equal, the arrangement is consecutive heads and consecutive tails.\
{H,H,H,T,T,T}\
The total number of arrangements of six coins = 6!\
Therefore, the required number of ways = $$\binom {6}{3}$$ \
The number of ways in which the number of tails is equal to the number of heads is \
$$\frac{6!}{3!3!}$$ = 20 ways

Now my doubt is that we have two ways of selecting either head or tails from each coin.. why dont we consider that?

• Doesn't matter how likely $H$ or $T$ are...you are told that there are three of each. That's all the information that matters.
– lulu
Feb 1 at 17:26
• Since the question is just asking for number of ways and not the probability or fraction of cases, we don't need to consider the total number of cases (two ways of selecting each coin).  If the question was phrased as "There are six coins, half of which are heads and half are tails. How many ways can they be arranged in a row?", would that make the solution easier to understand?  Or perhaps "There are six balls, half of which are black and half are red. How many ways can they be arranged in a row?", since coins are associated with probability. Feb 1 at 17:26

Yes, there are two possible orientations for each coin and you are already considering that - when you decide $$3$$ will be heads and $$3$$ will be tails.
The thing is - because the coins are identical, there's only one way to make this choice. Pick any three coins ($$1$$ way) and give them heads. The remaining get tails.
So, you can multiply by $$1$$ if you want to. It won't make any difference though.
You end up with two sets of coins , $$3$$ heads and $$3$$ tails. The number of ways to arrange them among themselves is
$$\frac{6!}{3! \times 3!}$$