Combinations and Permutations in coin tossing
Matthew Harrington 
I understand the formulae for combinations and permutations and that for the binomial distribution. However, I'm confused about their application to coin tossing.
Consider three tosses. Outcomes with two heads are HHT, HTH and THH. So, there are three and that's what you get if you use the formula for combinations.
With combinations "the order is not important". How does that apply in this case? In what way are these three outcomes different if the order is not important?
What are the 6 permutations?
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$\begingroup$Imagine that you have three balls numbered $1$, $2$, $3$ representing the flips on which heads may come up. (I'm using your scenario of three flips). You want to look at cases where you get two heads in three flips, so you are going to pick two of the balls. Picking first ball $1$ and then ball $3$ gives the same result (sequence of tosses) as first picking ball $3$ and then ball $1$ (both result in the outcome HTH).
Thus you are making an unordered selection of two of the three positions.
$\endgroup$ 5 $\begingroup$The six permutations exist only when you have three possible outcomes, not with the coin toss. If you have outcomes A, B and C, then the permutations are: ABC, ACB, BAC, BCA, CAB and CBA.
$\endgroup$ 1 $\begingroup$I marked paw88789's response as the answer because it helped me most among those here, but I did look in lots of other places for an explanation. I found a helpful one here:
It suggests thinking of 5 distinguishable coins and considering the number of ways you can arrange H1, H2, H2, T1, T2. This is 5!. Since there are in fact three indistinguishable heads and two indistinguishable tails the answer is 5!/(3!.2!).
It was the notion that order didn't matter that confused me, which might lead someone to think that HHT is the same as HTH is the same as THH, but as paw88789 says it's the order in which the heads are allocated to the positions heads occupy that doesn't matter.
$\endgroup$ $\begingroup$Coin toss series can be viewed, depending on what you want to know, as either "combination" or "permutation" but in all cases "with repetition" (meaning same side can occur again and again).
*Probably the best page that summarizes the Combination vs Premutation with or without Repetition *
If the question is "How many ways a series of R coin tosses (N=2 sides) can go? Of these, how many will have 2 Heads in the row?", you are looking for Permutation with Repetition where "HHT" is different outcome from "THH".
Permutation with Repetition is the simplest of them all:
N to the power of R.
Example:
3 tosses of 2-sided coin is 2 to power of 3 or 8 Permutations possible.
In these, "at-least-2 Heads in a row" permutations are: HHH, HHT, THH - 3. Probability of "at least 2 heads in a row" is 3/8th (0.375)
If the question is "If you throw a 2-sided coin (N=2), R times, how many times can you get at least 2 heads?", you are looking for Combination (order is not important) with Repetition where "HHT" and "THH" are same outcomes (combination).
Combination with Repetition formula is the most complicated (and annoying to remember): (R+N-1)! / R!(N-1)!
For 3 2-sided coin tosses (R=3, N=2), Combination with Repetition: (3+2-1)! / 3!(2-1)! = 24 / 6 = 4
These are (because order is not important): HHH, HHT, HTT, TTT
If you are looking for "at least 2 Heads", 2 options match: HHH and HHT (order not important).
Probability of you getting at least 2 heads is 2 outcomes / 4 Combinations (with Repetition) = 0.5. Right?
$\endgroup$ $\begingroup$Permutations count how many ways you can order $n$ distinct objects. So if you had a 3 sided object you tossed you'd get $3!$ patterns.
Here there are only 2 outcomes. So if you now label the 3 tosses as 1st, 2nd and 3rd, there are $2^3$ possible toss patterns of 3 tosses.
If you look at a restricted case, such as 2 heads one tail, then you can choose the positions of the 2 heads in $\binom{3}{2}$ ways. Which is the same as choosing the position of the single tail in $\binom{3}{1}$ ways. This is why your event "2 heads one tail" can happen in the number of ways you pointed out, given by the formula for combinations.
$\endgroup$ 1 $\begingroup$Lets say our aim is to find the probability of occurrence of two heads and one tail in three coin flips. To solve this lets start by naming the two heads and a tail in three coin flips. Lets name the heads as H-a and H-b. Lets name the tail as T. Now based on permutation we can find the arrangements of H-a, H-b and T in the three coin flip positions we have by computing 3p3 = 6. The actual permutations are listed below:
H-a H-b T H-b H-a T H-a T H-b H-b T H-a T H-a H-b T H-b H-a
But for our probability calculation we don't want to distinguish between H-a H-b and H-b H-a, so we divide 3p3 by 2p2 (2p2 gives the permutation of 2 items in 2 positions). So answer is 3p3 / 2p2 = 6/2 = 3.
So the probability of getting two heads and one tail is 3/8.
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