I don’t think these are good examples for ordered vs. unordered selections. Ordered selections are mainly used when you actually care about the order in which items are selected in a sequence. Also, honestly, I’d recommend learning the counting method instead of relying on this probability-based approach, because that can get confusing in more complex scenarios. Like if I have to attempt these, I will do something like -
1. 9 people & 2 pair of siblings
Total ways to select 2 people = 9C2 = 9x8/2 = 36
How many ways to select 2 siblings = 2
P(not siblings) = 1- P(siblings) = 1 - 2/36 = 17/18
2. 7 people & 2 pair of siblings & 1 pair of triplets
Total ways to select 2 people = 7C2 = 7x6/2 = 21
Total ways to select 2 siblings = 2 + 3C2 = 5
P(not siblings) = 1- P(siblings) = 1 - 5/21 = 16/21
Benefit of doing it this way is that you can always count the combinations which you are including or excluding, and hence you won't end up in the ordered or unordered trap which would be difficult for you to verify in the end when you are solving it under the clock.
uzerr
Thanks for the explanation. Is there a way of determining when to use ordered vs unordered selections?
When I try to do Q1 like Q2 (3 groups- 2 pairs siblings, 1 group of 5 with no siblings), I get 2 * 2/9*2/8 + 2 *5/9*8/8= 88/72 which is completely wrong
Knowing the answer is 17/18, I assume it would have to be 2 * 2/9*2/8 + 2 *5/9*6/8= 68/72. But I'm not sure what 6/8 would represent.I have read the threads on these problems, which give solutions which I can understand. But I still can't pinpoint when to use unordered vs ordered selections, and why using ordered pairs on Q1 didn't work.