Discrete pro. distribution: binomial We know that for a binomial distribution, when we want to know how many of the outcomes of an event has occurred rather than using a tree diagram, we can use selections, or combinations. For example, let a random variable X represents the number of heads after a coin is tossed three times, and we want to know the prob. of heads coming out once.
We would say,
Pr(X=1)= 3C1 times ... prob. of success times prob. of failure.
Because we know that there are three ways in which we could choose one head. From the tree diagram : HNN, NNH, NHN. H= heads, N= No heads.
My question is why is it correct to use combinations when it is clear that we don't use combinations for things where order matters. In here we can see that because these HNN, NNH, NHN are all different things containing the same element of one head, and two heads, it is clear that order does matter. Why can't we use permutations instead?
 A: Permutations count arrangements of distinct objects.  The elements of a sequence of heads and tails cannot be distinct if the sequence has length greater than two.
For instance, the number of permutations of the letters of the word COUNT, which has five distinct letters, is
$$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = P(5, 5)$$
and the number of three-letter permutations of the letters of the word COUNT is
$$5 \cdot 4 \cdot 3 = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1} = \frac{5!}{2!} = \frac{5!}{(5 - 3)!} = P(5, 2)$$
On the other hand, the number of distinguishable permutations of the letters of the word DISTRIBUTION, in which not all the letters are distinct, is
$$\binom{12}{3}\binom{9}{2}7! = \frac{12!}{3!9!} \cdot \frac{9!}{2!7!} \cdot 7! = \frac{12!}{3!2!}$$
since we must choose three of the twelve positions for the Is, two of the remaining seven positions for the Ts, and then arrange the seven distinct letters D, S, R, B, U, O, N in the remaining seven positions.  The factor of $3!$ in the denominator represents the number of ways we could permute the Is among themselves within a given arrangement without producing an arrangement which is distinguishable from the given arrangement; the factor of $2!$ in the denominator represents the number of ways we could permute the Ts among themselves within a given arrangement without producing an arrangement which is distinguishable from the given arrangement.
In your example, we use combinations since a sequence of heads and tails is completely determined by selecting the positions of the heads, as the remaining positions of the sequence must be filled by tails.
In general, in a binomial distribution problem, we define one of the outcomes to be a success and the other outcomes to be failures.  The probability of obtaining exactly $k$ successes in $n$ trials, each with probability $p$ of success is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ counts the number of ways those $k$ successes can occur in $n$ trials.  Notice that choosing which $k$ of the $n$ trials are successes completely determines the outcomes if there are exactly $k$ successes since the remaining $n - k$ trials must result in failures.
