Please help, I ALWAYS mess up on this type of math questions.

<p>First off, let me let you alllll know that I LOVE the math sections (I got a 680 math and 520 CR). Every time I get to one its a break from those deadly CR sections.</p>

<p>Nevertheless, I HATE it when I run into this type of problem:</p>

<p>the "how may possibilities are there?" questions.</p>

<p>For example:</p>

<p>"Dave is arranging 5 books on a shelf. If all five books must be arranged in one straight line, how many different orders ca he put the books in?"</p>


<p>May someone please show me the trick behind this type of questions?</p>

<p>ALSO: may any of you put up more questions of this type and explain how to get the answer?</p>


<p>B. 120</p>

<p>When you choose which to put first, there are 5 choices, then 4 choices for the second, 3 for the third and so on.</p>


<p>You have 5 open spaces for the first book. You have 4 open spaces for the second book. You have 3 open spaces for the third book. So on, so on.</p>

<p>From this you just do 5x4x3x2x1.</p>

<p>^thanks for the replies, may any of u (or anyone else) be willing to put up another example of this type of problem?</p>

<p>thanks again!</p>

<p>I'm not sure if you've learnt this, but the general way to tackle such questions is by using combinations and permutations. </p>

<p>Combinations refers to the number of ways you can take the items (not including order) while permutations refer to the number of ways you can <em>arrange</em> the items (including order). </p>

<p>In general, the formula is nCr (for number of combinations) and nPr (for number of permutations). n is the total choices you have, and r is the number you need.</p>

<p>So, since you have 5 choices of books, and you are going to arrange 5 of them, you need to use 5P5, which is 120.</p>

<p>I wrote some stuff on permutations and combinations here, you can go take a look.
<a href=""&gt;;/a&gt;&lt;/p>

<p>Permutation method
n P r is the number of permutation (ordered pairs) that you can have, given n options and r number necessary.</p>

<p>E.g. I have 7 bracelets. I decided to wear one on my wrist and one on my ankle? How many different get-ups do I have?
7 P 2 because wearing the pearl bracelet on the wrist and bead bracelet on ankle is different from wearing bead bracelet on wrist and pearl bracelet on ankle.</p>

<p>E.g. I want to choose two people to act as Romeo and Juliet. I have 10 cast members. How many different pairs of Romeo and Juliet can I have?
Answer: 10 P 2 (10 options, 2 needed)</p>

<p>E.g. I miss my 7 friends. I will go out with all 7 of them this week, one each day from Monday to Friday. How many different social calendars can I have?
Answer: 7 P 7 (7 options, 7 needed)</p>

<p>NOTE: You may also do these questions by the method that the previous posters have mentioned.</p>

<p>^i am lost</p>

<p>i dont understand how 5P5 means something that gives u 120</p>

<p>please help

<p>You probably haven't done P&C in school before, then. nPr just means n!/(n-r)! Anyhow, it may be useful to remember it.</p>

<p>5!/(5-5)! = 5!/0! = 5! = 5 x 4 x 3 x 2 x 1 = 120.</p>

<p>But, you don't need to memorise the formula as the calculator can do it for you. There should be a button on your calculator saying nPr or nCr. </p>

<p>Otherwise, here's an online calculator to work it out first: Combinations</a> and Permutations Calculator</p>

<p>This is an interesting tutorial on P&C: Combinations</a> and Permutations</p>