<p>How many different positive four-digit integers can be formed if the first digit must be 2, the last digit cannot be 0 and digits may be repeated? </p>
<p>:- :- plz how can i solve this type of questions ?? ?? plz give me a rule</p>
<p>This is a counting problem, and it would be a big help if you understand the concept and calculations of combinations and permutations.</p>
<p>The first digit must be a 2, so we don’t do anything there.</p>
<p>The second can be any of the 10 digits from 0-9, and we choose one of them, so the total number would be 1C10 = 10.</p>
<p>The third can be any of the 10 digits from 0-9 (due to repetition), and we choose one of them, so the total number would be 1C10 = 10.</p>
<p>The fourth can can be any of the 10 digits from 0-9 EXCEPT for 0, and we choose one of them, so the total number would be 1C9 = 9.</p>
<p>Now, we multiply them together to get 900.</p>
<p>If you want a rule, when doing problems like this, since we are basically looking for the number of permutations (order does matter), then just multiply how many possible digits/events in each space/placement. Like for instance, you can only have a 2 (1 digit) first, then you can have any of the 10 digits for the second and third digits, then you can have 9 possible digits last because you can’t have a zero. Multiplying them all gives you 900.</p>
<p>thnx !!! i got it :D<br>
but how can i deal with it if the numbers’ ( order does not matter ) combination ?</p>