In how many ways can 7 different objects be divided among 3 persons so that…

Concept: When n distinct objects are distributed among r persons (each object is independently assigned to any one of the r persons), the total number of distributions is rn.

The event "at least one person receives nothing" is the complement of a surjective distribution, where every person gets at least one object. Counting the empty-person cases directly, or subtracting the surjective count from the total, gives the same result.

Application:

1. Total unrestricted ways to give the 7 distinct objects to 3 persons = 37 = 2187.

2. Case — exactly two persons get nothing: all 7 objects go to a single person. Choose that one person: C(3,1) = 3 ways (only 1 way to hand every object to the chosen person, since every object goes to the same destination).

3. Case — exactly one person gets nothing: choose the empty person in C(3,1) = 3 ways; the 7 distinct objects are then split between the remaining two persons so that neither of them is empty. Splitting 7 distinct objects freely between 2 named persons gives 27 = 128 ways; removing the 2 extreme splits (all 7 objects to just one of the two) leaves 128 − 2 = 126 valid ways. So this case contributes 3 × 126 = 378.

4. Required count = (exactly one empty) + (exactly two empty) = 378 + 3 = 381.

Cross-check (inclusion–exclusion): the number of ways all three persons receive at least one object (a full surjection) is 37 − C(3,1)·27 + C(3,2)·17 = 2187 − 384 + 3 = 1806. Since all 7 objects must be placed somewhere, it is impossible for all three persons to be empty at once, so "at least one person empty" is exactly the same event as "one or two persons get nothing." Hence the required count = 2187 − 1806 = 381 — the same value, confirming the case-based total.

Result: The number of valid distributions is 381.