In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

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Solution

The correct option is A
16

$a + b + c + d = 7 a^{'} + b^{'} + c^{'} + d^{'} = 3$
Number of possible solutions = $^{6} C_{3} - 4 = 20 - 4 = 16$ .

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In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

The correct option is A 16 a+b+c+d=7a′+b′+c′+d′=3 Number of possible solutions =6C3–4=20–4=16.

The correct option is A
16

$a + b + c + d = 7 a^{'} + b^{'} + c^{'} + d^{'} = 3$
Number of possible solutions = $^{6} C_{3} - 4 = 20 - 4 = 16$ .