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Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter? $\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

Source of the problem
American Mathematics Competition

Combinatorics

7/10

Suggested Book

Enumetarive Combinatorics – ( Problem Solving Strategies ) by Arthur Engel

Connected Program at Cheenta

You could give it a thought first…are you sure you really need a hint ?

First can you try finding the number of ways in which we can choose the 2 shared elements ?  That would be 5C2 = 10 ways. Can you try completing this ?

All that remains now is to place the remaining 3 elements into the subsets. In how many ways can we do this ?

So basically, if we think about it a little, it is equivalent to the problem of placing 3 elements in 4 gaps. Evidently, this can be done in 4C3 = 4 ways.  Now, it’s pretty easy to arrive at the answer…could you do it by yourself ?

The final answer can be easily found out using the Multiplication Principle, which leads us to… 4 x 10 = 40 ways

And, we are done !

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