Consider the statistics of two sets of observations as follows:

If the variance of the combined set of these two observations is $$\frac{17}{9}$$, then the value of $$n$$ is equal to ________.

For the first set: $$n_1 = 10$$, $$\bar{x}_1 = 2$$, $$\sigma_1^2 = 2$$, so $$\sum x_i = 20$$ and $$\sum x_i^2 = n_1(\sigma_1^2 + \bar{x}_1^2) = 10(2+4) = 60$$.

For the second set: $$n_2 = n$$, $$\bar{x}_2 = 3$$, $$\sigma_2^2 = 1$$, so $$\sum y_j = 3n$$ and $$\sum y_j^2 = n(1+9) = 10n$$.

The combined mean is $$\bar{x} = \frac{20 + 3n}{10 + n}$$.

The combined variance is $$\sigma^2 = \frac{\sum x_i^2 + \sum y_j^2}{10+n} - \bar{x}^2 = \frac{60 + 10n}{10+n} - \left(\frac{20+3n}{10+n}\right)^2$$.

Setting this equal to $$\frac{17}{9}$$: $$\frac{(60+10n)(10+n) - (20+3n)^2}{(10+n)^2} = \frac{17}{9}$$.

Expanding the numerator: $$(60+10n)(10+n) = 10n^2 + 160n + 600$$ and $$(20+3n)^2 = 9n^2 + 120n + 400$$.

Numerator = $$n^2 + 40n + 200$$.

So $$9(n^2 + 40n + 200) = 17(10+n)^2 = 17(n^2 + 20n + 100)$$.

$$9n^2 + 360n + 1800 = 17n^2 + 340n + 1700$$, giving $$8n^2 - 20n - 100 = 0$$, i.e., $$2n^2 - 5n - 25 = 0$$.

$$n = \frac{5 \pm \sqrt{25 + 200}}{4} = \frac{5 \pm 15}{4}$$. Since $$n > 0$$, $$n = \frac{20}{4} = 5$$.

The answer is $$5$$.