<p>How many ways can seven books be arranged on a shelf if two are math books and must be kept together?</p>
<p>This is from Dr. Chung's. </p>
<p>Thanks</p>
<p>First of all, this is not a probability problem. It is a permutation based problem.</p>
<p>Solution:</p>
<p>think the 2 math books as one single book. So, there are 6 books to arrange on the shelf.</p>
<p>So, it can be kept in 6! ways. And the 2 math book can be arranged in 2! Ways.</p>
<p>So, the books can be arranged in 6! * 2! = 1440 ways.</p>
<p>I don’t have chung’s book. So, please can SOMEONE TELL ME IF I’m correct 
?</p>
<p>Okay, let’s say the books are named A-G, and A & B need to be kept together. There are twelve ways that A & B can be together since the first one can be in one of any of the first six positions and the book to go first can be switched, making a total of twelve permutations for the first too books. The remaining five books can occupy any leftover position, so their permutations are 5P5, or 5!, since you pick from 5 for the most left available position, you pick from 4 for the next most left available position, and so on. To get the total permutations, you multiply 12 x 5! to get 1,440.</p>
<p>By the way, a question like this will never appear on the SAT.</p>
<p>@johnstucky</p>
<p>Why is that?</p>
<p>Oh yeah, you’re correct. I made it way too complicated. I guess it could be on the SAT, but it seems a little too difficult compared to the other probability questions that I’ve see before.</p>
<p>So, I was correct . It won’t be in SAT 1 . But if you’re going to take SAT math level 2, You’ll need it.</p>
<p>Why wouldn’t the two books count as two different spots? </p>
<p>If the two books are placed together, there are only 5 books left. </p>
<p>So if you consider it as 1, </p>
<p>1<em>5</em>4<em>3</em>2*1 = 120 </p>
<p>But there’s 6 ways to arrange the two math books so 6*120 = 720.</p>
<p>Why isn’t this correct?</p>
<p>EDIT: My mistake, I didn’t consider the books switching spots so it would be 12 ways to arrange the books so,</p>
<p>12*120= 1440</p>
<p>There is some rules in permutation.</p>
<p>The basic rule is the probable ways that one can fill in r spots from n options is nPr . </p>
<p>You need to understand permutations and combinations plus probability.</p>
<p>If you do it in 1<em>1</em>2<em>3</em>4*5=120 in math tests you’ll get a big round zero. You can’t do permutations in this way. Sorry, this explanation is going to be big.</p>
<p>So. The answer isn’t correct because,</p>
<p>when you’re choosing 2 math books as 1, there are 6 books. And you can choose to place the book in places</p>
<p>1-2 , 2-3, 3-4, 4-5, 5-6, 6-7 and inverse ways as follows</p>
<p>7-6, 6-5…2-1 .</p>
<p>So, there are 12 places you can keep the books. And the remaining 5 places can be filled in from 5 available books. So, 5!.</p>
<p>What you’re doing is ignoring the reverse order. 7-6…2-1.</p>
<p>That’s why it is wrong.</p>
<p>If you’re making mistakes like this, then let me tell you, these aren’t silly mistakes. These are serious conceptual problems.</p>
<p>I caught my mistake in my last post.</p>
<p>But thanks anyway.</p>
<p>^ I got that after posting my response.</p>