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I am aware there is a question similar to mine here and here, however I'm asking because I have a more specific question regarding this issue. Some explanations I've gotten as to why we use combination is that the order doesn't matter. HTHT and HHTT is the same number of heads. But then wouldn't this mean that the "number" of ways we get two heads is just one? Obviously not because ${4\choose2}$ is not 1.

Furthermore, can someone explain what $\frac{n!}{(n-k)!}$ would mean in the context of the "4 trials, 2 heads" example? In permutations since order matters, doesn't this mean we count HHTT, HTHT, HTTH, and so on? But then when I add this up this comes to a total of 6, which is actually ${4\choose2}$!

user6246608
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3 Answers3

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Your confusion is on the meaning of "the order doesn't matter".

When we say that there are $4\choose 2$ ways to get two tails in 4 coin flips, it's because there are $HHTT, HTHT, HTTH,$ etc. There is a first coin flip, a second coin flip etc. and in this respect, the order matters as $HT$ is not considered the same as $TH$.

However, to count the amount of ways to get $2$ tails in $4$ flips, we count the number of ways to choose $2$ placeholders from a total of $4$ possible options. To get $TTHH$, we choose the first two placeholders. And here is where the order doesn't matter: if you choose the first placeholder, and then you choose the second, or if you choose the second and then the first, in the end you get $TTHH$. In other words you don't even have to choose them one by one, you can choose your two placeholders at once.

Arnaud Mortier
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    Hmmm I think I see what you mean. So it is not the HHTT vs HTHT that we are considering to "not matter," rather it is the, say, H1,H2,T3,T4 vs H2,H1,T3,T4 (where each number is the ith flip that happened) that the combination considers as the same thing? – user6246608 Nov 03 '19 at 23:42
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    @user6246608 Precisely. – Arnaud Mortier Nov 03 '19 at 23:53
  • Got it, thank you! I am curious though, then, is there any "meaning" per se to the number of permutations n!/(n-k)! in this scenario? Additionally, is a way to accurately phrase the combination: essentially how many different ways can two blue beads and red beads be arranged in a line? – user6246608 Nov 04 '19 at 00:09
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If we are taking no reference to order, then there is only one way to have two heads. This is because the total number of flips must add up to $4$. Likewise, the number of ways for no heads, one head, etc. is also one for the same reason.

In the example however, you seem to have a slight mistake in your mindset. We aren't exactly choosing the number of heads with order not mattering. This is a categorization problem. It's the four distinct flips that we are categorizing into one of two categories: Heads or Tails. What doesn't matter is the order that we pick which flip gets to fall under the category of Heads. Arnaud's answer explains this in a little more detail.

An example where this sort of "Permutation vs Combination" decision comes in is problems where we have a race with, say, $12$ contestants. If I were to want to know the number of ways the top $3$ could finish with respect to first, second, and third place, then this is a Permutation problem. However, if I only care about the number of ways to make a group of the best $3$ racers, then this is a Combination problem.

WaveX
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We are measuring the probability for: some arrangement of $r$ successes among $n$ trials, each with an identical and independent rate of success of $p$.

Now the probability for a particular arrangement of such is: $p^r~(1-p)^{n-r}$.

However, since the order of the successes does not matter, all such arrangements producing that outcome must be accounted among the measure.

Now the count for distinct ways to select places to put $r$ successes among $n$ trials, to form such an arrangement, is: $\binom nr$.

Therefore the probability we seek is :$$\dbinom nr~p^r~(1-p)^{n-r}$$

Graham Kemp
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  • 'since the order of the successes does not matter, all such arrangements producing that outcome must be accounted among the measure.' Sorry would you mind elaborating a bit on this please? I still can't figure out why we multiply $p^r~(1-p)^{n-r}$ with $\binom nr$, instead of multiplying it with permutation... – Constantly confused Feb 17 '20 at 21:18
  • Among the $2^4$ possible outcomes of four flips of a coin, the event of exactly two heads among four tosses is comprised of these $\tbinom 42$ outcomes :$$\rm{HHTT, HTHT, HTTH,THHT, THTH,TTHH}$$Because it does not matter to you in which order the heads and tails occur, you must count all possible orderings to evaluate the probability.$$\dbinom 42 \dfrac 1{2^4}$$ – Graham Kemp Feb 17 '20 at 22:11
  • By the way, here $\binom 42$ counts ways to select 2 from 4 positions, to place the two heads among the four flips. @DanielMak – Graham Kemp Feb 17 '20 at 22:15
  • Why doesn't the order in which the heads and tails occur matter...? I think this is my sticking point, because if it doesn't matter then wouldn't $HHTT$ and $TTHH$ count as one...? I am sorry if this sounds gibberish but I am really confused about the order... – Constantly confused Feb 17 '20 at 22:27
  • $\binom nr$ tells us how many ways to give a specific subset right? e.g. $\binom 32$ of ${A,B,C}$ would tell us how many subsets of size 2 there would be, and that would be $AB,BC,AC$, which is 3 – Constantly confused Feb 17 '20 at 22:33
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    Yes, and likewise $\binom 42$ counts subsets of size 2 of a set of size 4 (such as ${1,2,3,4}$). These six subsets are $${1,2},{1,3},{1,4},{2,3},{2,4},{3,4}$$Each of these subsets describes a distinct way to select placement of two heads among four flips - so each is a distinct outcome of the event "two heads among four flips". – Graham Kemp Feb 17 '20 at 23:13
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    Counting ways to do something when order is not important must account for how many orders there are to do that something. – Graham Kemp Feb 17 '20 at 23:19
  • @GrahamKemp. Amazing explanation! Thank you. Why the original formula of binomial distribution $x\sim Binomial(r,p)$, then $p(x=r) = p^r(1-p)^{n-r} $ has no combinatorial but when we say $p(x\ge r) = \sum_{k=7}^{n=10} \binom{n}{k} p^k(1-p)^{n-k}$? So I mean for one point it should be, $p(x=r) = \binom{n}{r} p^r(1-p)^{n-r}$ – Avv Apr 04 '21 at 16:25
  • The binomial pmf does include a binomial coefficient. – Graham Kemp Apr 04 '21 at 22:24
  • Now the probability for a particular arrangement of such is: $p^r~(1-p)^{n-r}$. How it can be calculated as product of probability in single experiment? Because $p$ is assigned in some space $\Omega$ ,however we are now in $\Omega ^n$ .How the probability of this two space related? – Meet Patel Sep 27 '23 at 00:52
  • The two spaces are related because $\Omega^n$ is constructed from $n$ independent instances of $\Omega$. @MeetPatel – Graham Kemp Sep 27 '23 at 02:56
  • @GrahamKemp Which probability measure we should consider on $\Omega ^n~$ while calculating such a things? i mean is there any natural measure? when $(\Omega,\mathcal A, P)$ is given . – Meet Patel Sep 27 '23 at 03:15