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How many DIFFERENT subsets of size 3 can be selected from a set of 7 elements?
A$\binom{7}{3} = 35$
B$P(7, 3) = 210$
C$3^7$
D$3 \cdot 7 = 21$
Answer & Solution
Correct answer: A. $\binom{7}{3} = 35$
1. Subset selection is UNORDERED: use COMBINATIONS not PERMUTATIONS.
2. $\binom{7}{3} = \dfrac{7!}{3!\,4!} = \dfrac{7 \cdot 6 \cdot 5}{6} = 35$.
3. Option B is the ORDERED arrangement count, $7 \cdot 6 \cdot 5 = 210$ (correct for ordered selection, wrong for subsets).
4. Options C, D are unrelated formulas.
_Source: Oscar Levin, "Discrete Mathematics: An Open Introduction", §1.3 (Combinations)._
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