Three distinct numbers are selected randomly from the set $$\{1, 2, 3, \ldots, 40\}$$. If the probability, that the selected numbers are in an increasing G.P. is $$\frac{m}{n}$$, $$\gcd(m, n) = 1$$, then $$m + n$$ is equal to _____.

The problem asks for the probability that three numbers chosen from $\{1, 2, \dots, 40\}$ form an increasing G.P.

1. Total Sample Space

The total number of ways to select 3 distinct numbers from 40 is:

$$n(S) = \binom{40}{3} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880$$

2. General Form of the G.P.

For the terms to be integers, they must follow the form $(kp^2, kpq, kq^2)$, where $\text{gcd}(p, q) = 1$ and $q > p$. The constraint is that the largest term $kq^2 \leq 40$.

3. Counting Favorable Cases

We iterate through possible values of $p$ and $q$:

• $p=1$: $q \in \{2, 3, 4, 5, 6\}$ yields $10 + 4 + 2 + 1 + 1 = 18$ cases.
• $p=2$: $q \in \{3, 5\}$ yields $4 + 1 = 5$ cases.
• $p=3$: $q \in \{4, 5\}$ yields $2 + 1 = 3$ cases.
• $p=4$: $q=5$ yields $1$ case.
• $p=5$: $q=6$ yields $1$ case.
• Total Favorable Cases:$$n(E) = 18 + 5 + 3 + 1 + 1 = 28$$

4. Final Probability and Result

The probability $P$ is given by:

$$P = \frac{28}{9880} = \frac{7}{2470} = \frac{m}{n}$$

Since $\text{gcd}(7, 2470) = 1$, we identify $m = 7$ and $n = 2470$.

$$m + n = 7 + 2470 = 2477$$