Something that is confusing me in binomial coefficient is that it is used when order does not matter; that’s why we divide on k!

But in the case of coins, let’s say we have 1 head of 3 tries. We might have 3 combinations:

HTT

THT

TTH

And the order does matter here, but we still use same Binomial coefficient to draw the Binomial Distribution

Hi kopachevsky,

You are talking about a classic example of **combination without replacement**. You are interested in the number of cases where ‘H’ appears once which is exactly the same as the number of cases where ‘T’ appear twice.

You don’t care whether ‘T’ appear the **second and third** positions or **third and second** positions. So it is the case in which the order does NOT matter. Furthermore, once the position in which ‘H’ is placed is determined, it cannot be replaced. So it is **combination without replacement**

I know it is confusing. Think carefully again and you get a sense. If not, put a question again.

Hi, Taekyo

I think I have some understanding here you mean T1-H-T2 is the same as T2-H-T1, it’s just hard to understand what difference is between T1 and T2 when you are dealing with coin tossing. I mean, you are just sequentially tossing coins, and it will be only T-H-T always.

Also in your example position of the H makes sense, so ordering partly involved maye you can propose some additional articles or videos explaining this.

Thanks

Why does the order matter? Each coin flip is independent.

For example, when we pick 3 numbers from 0 to 9 and saying that order does not matter, we’ll shrink down 123, 321 and 213 to only one pick 123, but when we do 3 flips of the coin we don’t narrow THH, HTH and HHT only to THH, only difference here is the position of T (order?)