how the following ways of selection differs from each other

Question

Suppose I want to select $2$ elements out of $6$ elements. Then I get $6C2 = 15$ combinations. Here $C$ represents the standard formula of ($N C R$). Now my question is what differences arise when we try to do the job in two tasks like first I choose $1$ element out of $6$ by $6C1$ and then $1$ among the remaining $5$ by $5C1$. The total combination becomes $6C1 \cdot 5C1 = 30$, which are not the same. I don't understand where I am wrong.

Answer 1

For the second case, the order of the first task and the task chosen matters. If we want to get rid of the order, divide it by $2!$.

$$\binom{6}{2}= \frac{6!}{4!2!}$$

Answer 2

By first method you calculate the cardinality of $\{\{i,j\}\mid (i,j)\in\{1,2,3,4,5,6\}^2\text{ and }i\neq j\}$.

By second method you calculate the cardinality of $\{(i,j)\mid (i,j)\in\{1,2,3,4,5,6\}^2\text{ and }i\neq j\}$.

Note that e.g. $\{3,4\}=\{4,3\}$ but $(3,4)\neq(4,3)$.

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Edit:

Applying the first method you find an answer on the question:

How many subsets of exactly $2$ elements has the set $\{1,2,3,4,5,6\}$?

Applying the second method you find an answer on the question:

How many ordered pairs $(i,j)$ are there if it is demanded that $i,j\in\{1,2,3,4,5,6\}$ and $i\neq j$?