Simple coin tossing question, confused on their answer

[mr_coffee]

Hello everyone. I"m just wondering why they solved the answer in this manner.

The question is:
A coin is tossed 4 times. Each time the result H for heads or T for tails is recorded. An outcome of HHTT means that heads were obtained on the first 2 tosses and tails on the second 2. Assume heads and tails are equally like on each toss.

Well they solved it this way:
There are 4 outcomes in which exactly one head can occur (since a string of one "T" and three "H"'s can have
the "T" in anyone of the string's four positions). So the probability of exactly one head is 4/2^4 = 1/4.

Okay I understand that there are 2^4 possible outcomes on 4 tosses, 2 chocies can happen, either a H or a T, so 2^4.

But Why did they say, since a string of one "T" and 3 "H"'s. If we are trying to find how many times you get exactly 1 head Why wouldn't they say the following:
since a string of one "H" and three "T"s can have the "H" in anyone of the strings for positions, such as HTTT, THTT, TTHT, TTTH, this shows its 4 ways, in which you will get exactly 1 head. I'm just confused on why they worded it that way or am I missing somthing?

Thanks

 

[quasar987]

It's probably just an error in the book.

 

[mr_coffee]

I thought so, but just wanted to make sure, the professor actually did the solution but either way it would come out to the same answer just making sure. thanks!