I tried using the Principle of Inclusion and Exclusion (PIE) but couldn't find a way through.
2026-04-04 07:55:01.1775289301
Find the number of 3 element subsets of the set {1, 2, …, 10}, in which the least element is 3 or the greatest element is 7.
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The number of subsets with least element $3$ is $\binom72$, for there are $7$ elements greater than $3$ and we must pick two from them to complete the subset ($3$ itself is fixed). Similarly, the number of subsets with greatest element $7$ is $\binom62$. Then the number of subsets satisfying both conditions is $3$, since the middle element can only be $4,5,6$.
By inclusion/exclusion, the final answer is $\binom72+\binom62-3=21+15-3=33$.