<p>Here are two counting problems to play with...</p>
<p>A school assigns each student a 4 character ID code. Each character can be any of the digits 0 -9 with no digits repeating. But the school is running out of codes.</p>
<ol>
<li><p>How many ADDITIONAL codes would be available if digits were allowed to be re-used multiple times? [This one is not so hard -- I've posted similar questions in the past...]</p></li>
<li><p>How many additional codes would be available if ONE digit were allowed to be repeated ONE time. So 1225 would be allowed but 1222 or 1551 would not. [I think this one is trickier...]</p></li>
</ol>
<ol>
<li><p>The original number of codes is 10<em>9</em>8<em>7 = 5040. If digits are re-used, the total number of codes is 10</em>10<em>10</em>10 = 10000. Therefore there are 10000 - 5040 = 4960 codes.</p></li>
<li><p>All we need to do is find the number of codes with exactly one digit being repeated once. There are three different digits we need, so 10C3 = 120 combinations. We need to multiply this by 3 to account for the one digit that is repeated (e.g. if we choose 1,2,3 our final string of digits could be {1,2,3,1}, {1,2,3,2}, {1,2,3,3}. 120<em>3 = 360. To arrange them into an ID code, we note that there are 4!/2! = 12 ways to arrange the four digits. Therefore there are 360</em>12 = 4320 additional codes.</p></li>
</ol>
<p>^ nice! </p>
<p>In the spirit of SAT prep, there is a solution to #2 that does not require nCr or nPr. (That’s my goal when I write these questions: how obnoxious can I make the problem while staying inside the SAT boundaries?)</p>
<p>I’ll post the solution later, but it uses nothing fancier than the counting principle…but it breaks the problem down into separate cases…</p>
<p>OK, so without nCr…</p>
<p>You can separate the problem into cases based on where you put the repeat digit. Call the repeat digit X. There are 6 ways to arrange the repeats:
X X _ _
X _ X _
X _ _ X
_ X _ X
_ X X _
_ _ X X</p>
<p>For each of these 6 ways, you have 10 choices for the digit that repeats, nine choices for the next, and 8 for the last…</p>
<p>6x10x9x8 = 4320</p>
<p>Not as elegant as rspence, but using less advanced math…</p>
<p>Ah, I see you’re writing your own SAT problems…nice.</p>
<p>If you want to make them obnoxious, put some bizarre-looking problem that has a really simple solution. Here’s an example I made up (it’s not SAT-level, but whatever)</p>
<ol>
<li>The numbers 3, 8, 12, 22, 36, 49, and 75 are written on a blackboard. Player A erases any two numbers and replaces them with the positive difference of the numbers. He continues this process until one number remains on the blackboard, the “final number.” Can the final number ever be zero? (Prove your answer)</li>
</ol>
<p>I write SAT problems all the time – I’m one of the old folks here at CC. I wrote “The New Math SAT Game Plan”. But, yes, I do think problem writing is good prep for the SAT, once you have reached a certain level. And I know that at least one of the other older (thoughnot as old as me) guys agrees with me: a while back, PWNtheSAT tried to drum up interest in a problem writing contest. I still think it’s a good idea. BTW, Pwn also wrote an SAT book. I don’t think it’s a coincidence that we both think problem WRITING gives you insight to the test. </p>
<p>As for your problem, I’ll play with it later. But another old guy rule: let students post first. (That way, we can pretend that we knew how to do it all along…)</p>
<p>Ah. Yeah, a problem writing contest looks pretty interesting. It’d have to be judged on how closely it resembles the SAT though…</p>