# How to know if its permutation or combination?

I have a question,

In how many ways can 6 tosses of a coin yield 2 heads and 4 tails?


Now, to me the question clearly seems to be of permutation as they have asked for the number of ways(they have mentioned no where that we have to choose.) But the actual solution tells that the question is solved by combination. Please help me out why is it a question of combination and not of permutation. Thanks

-
You are choosing, but in a somewhat subtle way: there are six tosses, which we can label 1, 2, 3, 4, 5, 6. You have to choose two of these to be heads, which can be done in ${}_6C_2$ ways, and you have to choose four of the four remaining tosses to be tails, which can be done in only ${}_4C_4=1$ way. Hence there are ${}_6C_2\cdot1$ ways. – Will Orrick Jun 15 '12 at 19:38

"Permutation" vs. "Combination" is not about "choosing" or "not choosing". It's about whether the order matters or not.

And there is also a difference between how you count and what you are counting.

Here, you are correct: the order "matters" in that you are looking at ordered sequences of six tosses, and you want to count in how many of these ordered sequence there are exactly 2 heads and 4 tails. (If the order of the tosses did not matter, then the answer would be "One: 2 heads and 4 tails").

But, how do you count them? The simplest way of doing this is to imagine you have six blank spots,

    __  __  __  __  __  __


and will place either an H or a T in each spot. You have two Hs and four Ts to put down, and you want to find out in how many ways you can do it. Once you put the two Hs down, the Ts will be forced, so you just need to find in how many ways you can put the two Hs. It doesn't matter which H you put down first. The only thing that matters is where the two Hs end up, not which one went first. So to count this we will use combinations (because order does not matter). And so you need to find the number of ways in which you can choose two of the six spots, which is where the Hs will go. Combinations, because the order in which the Hs are placed does not matter: deciding the first and then the third tosses will be head amounts to the same thing as deciding that the third and the first tosses will be heads.

-

If order does not matter, it is a combinations problem. If order does matter, it is a permutations problem. With your example, it does not seem the order of the heads or tails occurance matters, so combinations is the way to go here.

-
"Obviously permutations would yield less ways than combinations...." How so? It is always the case that $nPr\geq nCr$, so that can't be what you mean. – Cameron Buie Jun 15 '12 at 17:09
Yes, you are certainly correct - $nPr \ge nCr$ for for all such outcomes. Let me think about how to explain it in another way other than with 'order'. – Joe Jun 15 '12 at 17:23

Are you familiar with "distinguishable permutations"? The idea is the same as with permutations, with the exception that there may be no ready way to tell the difference between some of the objects being arranged.

For example, how many ways can we arrange the letters of TENNESSEE? Well, the answer to that is clearly $9!$ since there are $9$ letters, but not all of those arrangements will be distinct, since we have repeats. Given a particular arrangement of the letters, how many ways can we rearrange just the E's so that the word looks the same? $4!$, since there are $4$ of them. Likewise, there are $2!$ ways to rearrange just the N's, or to rearrange just the S's. Thus, there are $\dfrac{9!}{4!2!2!}$ distinguishable permutations of the given letters.

More generally, given objects $O_1,...,O_k$ (where $O_i,O_j$ are distinguishable for $i\neq j$), if we have $n_j$ of the object $O_j$ for each $1\leq j\leq k$, and cannot distinguish between two $O_j$'s, then if $N=n_1+...+n_k$, there are $\dfrac{N!}{n_1!\cdots n_k!}$ distinguishable permutations of that collection of objects.

Applied to this particular situation, we have $O_1:=H$, $O_2:=T$, $n_1=2$, $n_2=4$, and so the number of ways to arrange them is $\dfrac{6!}{2!4!}$. In this case, that happens to be the same as $\binom{6}{2}=\binom{6}{4}$, so even though order matters, in a sense, we needn't consider the order as making a difference, here, because of lack of distinguishability, and the fact that we are only considering $2$ possible outcomes.

If there were $3$ or more possible outcomes, then we shouldn't try to use combinations. Say we want to know how many ways we can roll a die $10$ times and get two $1$'s, five $3$'s, and three $4$'s. Applying the distinguishable permutations formula, the answer is simply $\dfrac{10!}{2!5!3!}$.

-

In this question, we are tossing the same coin. Consider an outcome where we get $2$ heads and $4$ tails, say $$HTTTHT$$ Since you are tossing the same coin, you cannot distinguish between the two heads. Similarly, you cannot distinguish between the four tails. Hence, there is only one way to get the outcome $HTTTHT$ in this order.

On the other hand, if we were to toss $6$ different colored coins to obtain $2$ heads and $4$ tails, then the outcome $$\color{red}{H}\color{blue}{T}\color{green}{T}\color{orange}{T}\color{black}{H}\color{magenta}{T}$$ is different from $$\color{blue}{H}\color{magenta}{T}\color{red}{T}\color{black}{T}\color{orange}{H}\color{green}{T}$$ even though the number of heads and tails and the order in which the heads and tails occur are the same.

We will now count the number of ways of getting the outcome $HTTTHT$ when the coins are colored differently. The first head can be obtained from any of the $6$ differently colored coins. Hence, the first position has $6$ options. Once the first position is fixed as a head, the second position i.e. the tail, can be obtained from any of the $5$ remaining coins. Fixing this, the third position i.e. again a tail, can be obtained from any of the remaining $4$ coins and so on. Hence, the total number of ways of getting $4$ heads and $2$ tails from $6$ differently colored coins in the order $HTTTHT$ is $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6! = 720$.

Now lets get back to the original problem i.e. tossing the same coin to get $2$ heads and $4$ tails.

To look at all possible outcomes, which give us $2$ heads and $4$ tails, by the tossing of the same coin, note that all possible outcomes resulting in $2$ heads and $4$ tails by tossing the same coin $6$ times can be obtained by permuting the outcome $HTTTHT$. The total number of permutations if we do blindly is $6!$.

However, note that for instance, if the two heads were to swap its place, keeping the tails fixed and since we are tossing the same coin, we cannot distinguish between them. Hence, to take into account these are two heads from the same coin, we need to divide by $2!$.

Similarly, if all the four tails were to permute among themselves without changing the position of the two heads, we cannot distinguish this outcome from the earlier one we had. Hence, to take into account these are four tails from the same coin, we need to divide by $4!$.

Hence, the total number of ways to get $2$ heads and $4$ tails is $$\dfrac{6!}{4! 2!} = \dbinom{6}{2} = \dbinom{6}{4}$$

-

Basically just as Arturo Magidin mentioned, if the order matters - permutation, if order doesn't matter - combination.

So in this case, let's focus on the two head, let's assume they are completely distinct heads: H1 and H2. And let's assume we get two heads in the first two toss and rest are tails, does it matter if it's H1 H2 T T T or H2 H1 T T T?

If it doesn't matter, combination. If it matters, permutation.

-