See image β AITS & Test Series Chemistry Question
Question
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Answer: A
π‘ Solution & Explanation
Let Λ Λ Λ r x yj zk ο½ ο« ο« ο² x + y + 4z ο£ 14; x, y, z ο N Number of triplets (x, y, z) = 1 + 6C4 + 10C8 = 61.
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