are to be chosen so that their union is 
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?
[/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
[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="4.0"]7/10
[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" open="off"]Enumetarive Combinatorics - ( Problem Solving Strategies ) by Arthur Engel
[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||117px|||" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]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 ?
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]All that remains now is to place the remaining 3 elements into the subsets. In how many ways can we do this ?
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]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 ?
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]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 !
[/et_pb_tab][/et_pb_tabs][et_pb_blurb title="Math Olympiad Program" url="https://cheenta.com/matholympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/matholympiad/" link_option_url_new_window="on"]Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
