Three distinct numbers are selected randomly from the set 1,2,3,,40.

Question

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 $\dfrac{m}{n}$, $\gcd(m,n)=1$, then $m+n$ is equal to \_\_\_\_.

Answer: 4949

Accepted Answer

Counting increasing G.P. triples $(a,ar,ar^2) \le 40$ by common ratio $r$ (integer $\ge 2$; $r=1$ excluded as terms must be distinct):

- $r=2$: last triplet $(10,20,40)$; 10 triples
- $r=3$: last triplet $(4,12,36)$; 4 triples
- $r=4$: last triplet $(2,8,32)$; 2 triples
- $r=5$: last triplet $(1,5,25)$; 1 triple
- $r=6$: last triplet $(1,6,36)$; 1 triple

Total favourable $=10+4+2+1+1=18$.

Total ways $={}^{40}C_3=9880$.

$$P = \frac{18}{9880}=\frac{9}{4940}$$

So $m=9,\ n=4940 \Rightarrow m+n=4949$.

Solutions · 1