Henry Ernest Dudeney/Puzzles and Curious Problems/47 - The Quarrelsome Children/Solution

by : $47$

 * The Quarrelsome Children

Solution

 * $6$

Proof
Let $A$ and $B$ denote the sets of the children of the man and the widow respectively.

Then
 * $A \cap B$ denotes the set of the children of both of them
 * $A \cup B$ denotes the set of all the children together.

From the Inclusion-Exclusion Principle:
 * $\card {A \cup B} = \card A + \card B - \card {A \cap B}$

where $\card X$ denotes the cardinality of a set $X$.

We have been given that:
 * $\card {A \cup B} = 12$
 * $\card A = \card B = 9$

which leads to:
 * $\card {A \cap B} = 9 + 9 - 12 = 6$