Question ID: #607
Three distinct numbers are selected randomly from the set $\{1, 2, 3, \dots, 40\}$. If the probability that the selected numbers are in an increasing G.P. is $\frac{m}{n}$ where $\gcd(m,n)=1$, then $m+n$ is equal to:
Solution:
Total ways to select 3 distinct numbers from 40 is:
$$ n(S) = {}^{40}C_3 = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880 $$
Let the numbers be $a, ar, ar^2$ in increasing order. The common ratio $r$ can be an integer or a rational number.
The condition is $ar^2 \le 40$.
* If $r=2$: $4a \le 40 \Rightarrow a \le 10$ (10 GPs: 1,2,4 to 10,20,40)
* If $r=3$: $9a \le 40 \Rightarrow a \le 4$ (4 GPs)
* If $r=4$: $16a \le 40 \Rightarrow a \le 2$ (2 GPs)
* If $r=5$: $25a \le 40 \Rightarrow a \le 1$ (1 GP)
* If $r=6$: $36a \le 40 \Rightarrow a \le 1$ (1 GP)
Total integer GPs = $10 + 4 + 2 + 1 + 1 = 18$.
$$ P(E) = \frac{18}{9880} = \frac{9}{4940} $$
Here, $m=9, n=4940$.
$$ m+n = 9 + 4940 = 4949 $$
Ans. (4949)
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