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All the non-ordered pairs of coprimes within $\{2,3,4,5,6,7,8,9\}$ are listed below:

$\begin{array} {lcccccccl} 2: & 3 & \color{#D61F06}{\not{4}} & 5 & \color{#D61F06}{\not{6}} & 7 & \color{#D61F06}{\not{8}} & 9 &\Rightarrow 4 \\ 3: & & 4 & 5 & \color{#D61F06}{\not{6}} & 7 & 8 & \color{#D61F06}{\not{9}} &\Rightarrow 4 \\ 4: & & & 5 & \color{#D61F06}{\not{6}} & 7 & \color{#D61F06}{\not{8}} & 9 & \Rightarrow 3 \\ 5: & & & & 6 & 7 & 8 & 9 & \Rightarrow 4 \\ 6: & & & & & 7 & \color{#D61F06}{\not{8}} & \color{#D61F06}{\not{9}} & \Rightarrow 1 \\ 7: & & & & & & 8 & 9 & \Rightarrow 2 \\ 8: & & & & & & & 9 & \Rightarrow 1 \\ \end{array}$

There are altogether $4+4+3+4+1+2+1 = \boxed{19}$ coprimes.

$\large \color{#3D99F6}{\underline{Alternative \space Solution}}$

There is another solution using Euler's totient function, which I have just learnt. Thanks to Abhisek Mohanty for asking the right question.

Euler's totient function $\varphi (n) = n \displaystyle \prod_{p|n} {\left(1-\frac{1}{p}\right)}$ , where $p$ is a prime, gives the number of positive coprimes of a positive integer $n$ less than $n$ .

Since the sequence starts with $2$ instead of $1$ , the number of coprimes of $n$ has to less $-1$ . And the number of unordered pairs of coprimes is given by:

$\begin{aligned} N & = \displaystyle \sum_{n=2}^9 {(\varphi (n)-1)} \\ & = \sum_{n=2}^9 { \left[ n \prod_{p|n} {\left(1-\frac{1}{p} \right)} \right]} - \sum_{n=2}^9 {1} \\ & = 2\left(1-\frac{1}{2} \right) + 3\left(1-\frac{1}{3} \right) + 4\left(1-\frac{1}{2} \right) + 5\left(1-\frac{1}{5} \right)\\ & \quad + 6\left(1-\frac{1}{2} \right)\left(1-\frac{1}{3} \right)+7\left(1-\frac{1}{7} \right) + 8\left(1-\frac{1}{2} \right) \\ & \quad + 9\left(1-\frac{1}{3} \right) - 8 \\ & = (2+4+8)\left(\frac{1}{2}\right) +(3+9)\left(\frac{2}{3}\right) + 5\left(\frac{4}{5}\right) +6\left(\frac{1}{2}\right)\left(\frac{2}{3}\right) \\ & \quad +7(\left(\frac{6}{7}\right) - 8 \\ & = 7+8+4+2+6-8 \\ & = \boxed{19} \end{aligned}$

In set (2, 3, 4, 5, 6, 7, 8, 9), there are totally 8C2= 28 non- ordered pairs can be formed. In set (2, 4, 6, 8), there are 4C2= 6 non- ordered pairs can be formed. And, in set ( 3, 6, 9), there are 3C2= 3 non- ordered pairs can be formed. So, non- ordered pairs of co- primes can be formed= 28- 6- 3= 19.

There are 7 integers greater than 2.4 out of them are co-prime with 2 (3,5,7,9)

6 integers are greater than 3,4 of them are co-prime with 3 (4,5,7,9)

5 integers are greater than 4,3 of them are co-prime with 4 (5,7,9)

All the 4 integers greater than 5 are co-prime with 5.

In case of 6,only 7 is co-prime with it.

8 and 9 are co-prime with 7.

8 is co-prime with 9 .

So the total is $4+4+3+4+1+2+1=19$