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There are 7 gap areas b/w 7 rings, so select 1 gap area and place 2 rings = 7C1*2!
Permutation for remaining 7 rings in a circle = 6!

Favorable outcomes = 7C1*2!*6!

Total permutations for 9 rings in a circle = 8!

Probability = \(\frac{7*2*6!}{8!} = \frac{2}{8} = \frac{1}{4}\)

IMO: D
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I've got a different way to solve this which somehow seems easier for me:

- Consider the keys as a set = [1,2,3,4,5,6,7,8,9] where 8 and 9 are the new added keys
- Problem asks for probability of choosing 2 adjacent keys (regardless of whether they are the 2 new added ones or any other 2)
- From 9 keys, there are 36 possible ways of choosing 2 keys at random: n!/r!(n-r)! = 9!/2!7! = 36
- Calculate how many ways we can pick 2 adjacent keys: 1,2 - 2,3 - 3,4 - ... - 1,9 and don't forget 9,1 since keys are in a ring - Total ways 9
- Probability of choosing 2 adjacent keys will be the number of possible picks out of total: 9/36 = 1/4
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