[/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_accordion_item][/et_pb_accordion][et_pb_text _builder_version=”4.0.9″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”48px||48px” custom_padding=”20px|20px|0px|20px||” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]
Start with hints
[/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 , since all of the valid 4-digit number will always be greater than . [/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 or .[/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 such numbers. [/et_pb_tab][et_pb_tab title=”HINT 6″ _builder_version=”4.1″]When the leading digit is then we have one 2, one 1 and one 0 then we can arrange them in 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″]
Similar Problems
[/et_pb_text][et_pb_post_nav in_same_term=”off” _builder_version=”4.0.9″][/et_pb_post_nav][et_pb_divider _builder_version=”3.22.4″ background_color=”#0c71c3″][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″ saved_tabs=”all”][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” _builder_version=”4.2.2″ button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” hover_enabled=”0″ custom_button_one=”on” button_one_bg_color=”#ffffff” button_one_bg_enable_color=”on” button_one_border_color=”#ffffff” button_one_text_color=”#44580e” button_one_border_radius=”5px” background_color=”#00457a” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png”]AMC – AIME – USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
[/et_pb_fullwidth_header][/et_pb_section]
, since all of the valid 4-digit number will always be greater than 
. [/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 
or 
.[/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 
such numbers. [/et_pb_tab][et_pb_tab title=”HINT 6″ _builder_version=”4.1″]When the leading digit is 
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″]