Using 5C2 to calculate number of ways to get 2 heads in 5 coin tosses

[Darkmisc]

Homework Statement

How many ways can you get two heads in five coin tosses

Relevant Equations

5C2 = 10

Hi everyone

I'm looking at binomial distributions and one of the examples uses 5C2 to calculate the number of ways you can get two heads from five coin tosses. I don't question that the formula is correct, I just find it hard to picture what the numbers represent.

The numbers make sense to me in the context of 5 choose 2. For example, if you are choosing a team of two from five people, 5 represents the number of choices for the first spot. Four represents the number of choices for the second spot. You then divide the permutations by two to avoid double counting.

What do the numbers mean in the case of five coin flips? I'm picturing it more as a case of 2 choose 5 (even though that's probably not a thing). You can choose between heads or tails and have five flips to fill.

If I were to represent the coin flip situation as 5x4/2, what would the 5, 4 and 2 mean?Thanks

 

[PeroK]

HHTTT
HTHTT
HTTHT
HTTTH
THHTT
THTHT
THTTH
TTHHT
TTHTH
TTTHH

 

[Darkmisc]

Actually, I think I get it now. Would this be a good way to explain it?

If you look at the permutations, a head can occur on any of the five flips. Once you know a head has occurred at a given flip, there are only four other possible places for the second head to occur. The permutations need to be divided by two to avoid double counting. Thus, 5x4/2.

 

[PeroK]

Actually, I think I get it now. Would this be a good way to explain it?

If you look at the permutations, a head can occur on any of the five flips. Once you know a head has occurred at a given flip, there are only four other possible places for the second head to occur. The permutations need to be divided by two to avoid double counting. Thus, 5x4/2.

That's the idea. You could code those ten combinations (note these are not permutations) by specifying the two places where the heads are. So, HHTTT becomes 12; and TTHTH becomes 35 etc.

And now this is equivalent to drawing two numbers from a pot of five numbers (without replacement) and where the order you draw the numbers does not matter. I.e. drawing number 1 then number 2 is the same as drawing number 2 then number 1. And that is more explicity $\frac{5 \times 4}{2}$.

 

[FactChecker]

If I were to represent the coin flip situation as 5x4/2, what would the 5, 4 and 2 mean?

I would not try to understand it that way. I would understand it in the form 5!/(2!3!),
The 5! is the number of ways to order 5 objects if they are all different.
The 2! is the number of ways that the two heads have been over-counted. Swapping the two heads should not be counted twice, so divide by 2!.
The 3! is the number of ways that the three tails have been over-counted. Swapping any of them should not be counted again, so divide by 3!.