<p>I'm having trouble with this problem from one of my SAT math books:</p>
<p>A five digit number is to be formed using each of the digits 1,2,3,4, and 5 exactly once. How many such numbers are there in which the digits 1 and 2 are next to each other?</p>
<p>(A) 36
(B) 48
(C) 60
(D) 72
(E) 120</p>
<p>Here's what I tried: There are 5 ways to place the 1. Once the 1 is placed then the 2 is determined. Then there are 3 ways to place the 3, 2 ways to place the 4, and 1 way to place the 5.</p>
<p>So I get 5<em>1</em>3<em>2</em>1=30. But the book says the answer is (B). What am I doing wrong?</p>
<p>Thanks for your help!!</p>
<p>Let * be a number, 3,4,5</p>
<p>The possible cases are</p>
<p>12***
<em>12</em>*
<strong>12*
*12
21</strong>
<em>21</em>
*<em>21</em>
***21</p>
<p>The number of times we can arrange 3,4,5 is 3! = 3x2x1=6</p>
<p>From above, there are 8 possible arrangements, so the answer is 8(6) = 48</p>
<p>All we want are the possible ways 1 and 2 can be placed in a 5 digit number, regardless of where the other 3 numbers are.</p>
<p>You could also consider 1-2 as a single digit in a four digit number. A four digit number has 4! permutations; since same logic applies to 2-1, total number of acceptable permutations is 2*4! = 48</p>
<p>Just know that the possible arrangements for the remaining numbers = 3<em>2</em>1 = 6.</p>
<p>With that in mind, just consider the different positions for 12 and 21 in the 5 number series, and that has already been explained in a post before; it’s 8.</p>
<p>8*6 = 48</p>
<p>this has cleared things up, thanks guys.</p>
<p>also, re-reading the solution in the book I can see what I was doing wrong…</p>