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# Combinatorics – AMC 10A 2008 Problem 23 Sequential Hints

AMC 10A 2008, Problem 23 needed a clever trick of set theory and combinations. See the solution with sequential hints for a subset theory-based problem

# Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]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? $\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.0″ open=”on”]American Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]

Combinatorics

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7/10

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Enumetarive Combinatorics – ( Problem Solving Strategies ) by Arthur Engel

# Connected Program at Cheenta

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”4.0″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” custom_padding=”||153px|25px||”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]You could give it a thought first…are you sure you really need a hint ?

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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 ?

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

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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 ?

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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 !