If 6 numbers are chosen at random from the numbers between 1-9, then how many times is the probability of choosing 2,3,4,5,6,7 in any order compared to the probability of choosing 4,5,6,2,3,7 in that order?

is the answer 720?

I don’t actually know the official answer.

In the question, let us consider the first case as event-1 (E1) and the second case as event-2 (E2).

As we can see, E2 can be achieved in only 1 way. i.e. First pick the number 4, next pick ethe number 5, next pick 6 and so on… So the probability of E2 will be \frac{number-of-ways-E2-can-occur}{number-of-total-occurances-of-all-events}. Let’s assume the denominator is T (we don’t need the actual value of T as we just need to compare). So P(E2)= 1/T.

Similarly, for E1 to occur → First we can pick any of the 6 numbers, next we can pick any of the remaining 5 numbers, and so on… So the number of ways E1 can occur is 6.5.4.3.2.1 (which is 720). Now P(E1)= 720/T.

Comparing the two, P1 is 720 times of P2.

Thank you, that’s how I solved it too and got 720!