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# Permutation and basic counting principle AMC 8 2012, problem 10

Try this beautiful problem from AMC 8. It involves permutation and basic counting principles. We provide sequential hints so that you can try the problem.

# Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.1″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?[/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=”4.1″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.1″]American Mathematical Contest 2012, AMC 8 Problem 10[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.1″ open=”off”]Permutation and basic counting principle[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.1″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.1″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.1″]For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$, since all of the valid 4-digit number will always be greater than $1000$.  [/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.1″]The best way to solve this problem is by using casework. Now think what are the cases?  [/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.1″]It has two cases , as there can be only two leading digits, namely $1$ or $2$.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.1″]We know  that number of ways of arranging ‘n’ items, out of which ‘p’ are alike, ‘q’ are alike and ‘r’ are alike given that p + q + r = n Number of ways of distributing ‘n’ distinct items, in groups of size ‘p’, ‘q’ and ‘r’ given that p + q + r = n . . Now try to calculate the the two cases .[/et_pb_tab][et_pb_tab title=”HINT 5″ _builder_version=”4.1″]CASE 1: As 2012 consits of two 2’s , one 1, 0 so if we set 1 as the leading digit then we have two twos and one 0 such numbers then we have  $\frac{3!}{2!1!} \implies 3$ such numbers.  [/et_pb_tab][et_pb_tab title=”HINT 6″ _builder_version=”4.1″]When the leading digit is $2$ then we have one 2, one 1 and one 0 then we can arrange them in  $3! \implies 6$ ways and as such we have 6 such numbers.[/et_pb_tab][et_pb_tab title=”HINT 7″ _builder_version=”4.1″]By addition principle we find that there are  9 such numbers.[/et_pb_tab][/et_pb_tabs][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″]