Permutation and CombinationMedium
Question
A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is
Options
A.18
B.36
C.39
D.72
Solution
Sol. [Key Concept: Distributing $4$ distinct books into $3$ distinct bags with none empty is an onto count $3!\ S(4,3) = 6 \times 6 = 36$.]
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