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Bunuel
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Very good quality question simple but easy to miss under exam condition because of time pressure.
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I wanted to clarify if this arrangement is possible too?

2 Polished Emeralds, 2 Polished Amethysts + P/UP Amethyst + P/UP Garnet

As the ques mentions the same number and finish of amethysts must be selected too (same as emeralds) :- minimum that no. of A are selected as E, but there's no rule that it has to be maximum too.
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harshitasinghal
I wanted to clarify if this arrangement is possible too?

2 Polished Emeralds, 2 Polished Amethysts + P/UP Amethyst + P/UP Garnet

As the ques mentions the same number and finish of amethysts must be selected too (same as emeralds) :- minimum that no. of A are selected as E, but there's no rule that it has to be maximum too.

That arrangement is not possible. The rule says:

Quote:
If more than one emerald is selected, the same number of amethysts, with the same finish, must also be selected.

This means the count must match exactly, not “at least.” So with 2 polished emeralds, you must have exactly 2 polished amethysts, no more and no fewer. Adding any extra amethyst (polished or unpolished) would break the 1-to-1 correspondence required by the rule.
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I like the solution - it’s helpful.
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