(a)2439. (b)4096. (c)4903. (d)4904. (e)5416
[/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" open="on" _builder_version="4.0"]American Mathematical Contest 2006 10 A Problem 21[/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" _builder_version="4.0" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills in Pre College MathematicsExcursion Of Mathematics
[/et_pb_accordion_item][/et_pb_accordion][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" min_height="148px" custom_padding="||24px|||"][et_pb_tab title="Hint 0" _builder_version="4.0"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]Step 1.After having a look into this problem you can see our main aim here is to find out the total number of 4 digit positive numbers with at least one 2 or one 3 in it. Now we can do this by subtracting the number of 4 digit positive number which do not have any 2's or 3's from the total number of 4 digit positive numbers. Its easy, give it a try.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Step 2.Now lets' proceed forward and try to find out the total number of 4 digit positive integers . We can use all digit from 0 to 9 as we do not have any restriction . Now to make a table with this
Step 3. Now for the 2nd part. So try to find out the total number of 4 digit positive integers which don't have any 2's and 3's in it . So now in this case we have 8 digits(2 and 3 are excluded). Now to make a table with this
Step 4.Now when you have found the number of positive 4 digit numbers in both the cases now
(total number of 4 digit positive integers)-( total number of 4 digit positive integers which don't have any 2's and 3's in it )=(total number of 4 digit positive numbers with at least one 2 or one 3 in it). Very close to the solution just about to crack it !!!.
[/et_pb_tab][et_pb_tab title="Hint 5" _builder_version="4.0"]Total number of 4 digit positive integers(Table 1) = 9000Total number of 4 digit positive integers which don't have any 2's and 3's in it(Table 2) = 3584Total number of 4 digit positive numbers with at least one 2 or one 3 in it = 5416 that's your required answer.
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