PSAT Math Problem

Test Preparation SAT Preparation
1

<p>(From the 2007 Saturday PSAT, section 4)
[ul]
[<em>] Each digit in the integer is 3, 4, 5, 6, or 7
[</em>] Exactly 2 digits in the integer are the same.
[li] Any 2 adjacent digits in the integer are different[/ul]</p>[/li]
<p>How many positive 3-digit integers satisfy all three conditions?</p>

<p>I really hate permutations/ combinations/ counting. Do I have to use the nCr or nPr formulas? This problem is kinda like this <a href="http://talk.collegeconfidential.com/sat-preparation/1287354-math-question-help.html#post13856169%5B/url%5D"&gt;http://talk.collegeconfidential.com/sat-preparation/1287354-math-question-help.html#post13856169&lt;/a&gt;&lt;/p>

<p>The correct answer is 20. </p>

<p>My work:
4! = 24
(Because the two digits that have to be the same count as one)</p>

<p>That's as far as I got.</p>

<p>You can get 20 by 5!/2/3 but that doesn't make sense.</p>

2

<p>Since two digits have to be the same but can’t be adjacent, you know it’s the two outer digits that are the same. And you have 5 choices for those outer digits, then 4 choices for the inner digit. 5 x 4 = 20.</p>

<p>This is a CLASSIC sat counting problem in that it is tricky looking but requires only the counting principle to solve.</p>

3

<p>Just know that the two same integers have to be the first and last digit, and you should be able to figure it out from there.</p>