One math question from BB 2nd edition!

Test Preparation SAT Preparation
1

<p>This is from test 4, section 6, question 18.</p>

<p>I am terrible at combinations/permutations...</p>

<ol>
<li>If the 5 cards shown above are placed in a row that that (card C) is never at either end, how many different arrangements are possible?</li>
</ol>

<p>The card arrangement shows 5 different cards, with a card I called "C" in the middle.
So it's like this:
A B C D E</p>

<p>The answer is 72. Can someone explain to me how to do these?
I've always sucked at probability/permutations/combination/whatever problems.
Any help is appreciated!</p>

2

<p>(1) For the card at the left end there are 4 choices because C is excluded.
(2) For the card at the right end there are now 3 choices since one of the choices was used up above, and because C is excluded.
(3) For the second card (from the left) there are 3 choices because 2 of the cards are already placed (left end and right end). Note that C is allowed.
(4) For the third card (from the left) there are 2 choices because 3 of the cards have already been placed. Note that C is allowed.
(5) For the fourth card (from the left) there is one choice left.</p>

<p>So multiply the possibilities for each of the 5 cards: 4<em>3</em>3<em>2</em>1 = 72.
For the second card there are 4 choices because one of the cards was already placed. C is allowed</p>

3

<p>Here’s an easy way to visualize it: Without card C, there are 4 cards, and 4! ways to arrange them. For each of those permutations, C can be placed between two cards in 3 ways. So, 3X4! = 72.</p>

<p>So, for cards W|X|Y|Z - the three bars are the insertion points for C.</p>

4

<p>It’s so much simpler like this, Mikeypz.
Let’s put the five cards in place as following:
A - B - C - D - E
These are the five different positions possible right? Now, one of the cards, has to be in the middle, OK? Alright, now let’s take a look at each position one by one.</p>

<p>(A), you were told a ONE card has to be in the middle. Is A the middle? No. That means there are only four possible cards that can be put in position A. Good, that’s our first number, 4.</p>

<p>(B), ah, this is the first one of the three middle positions. Let’s assume we are not going to put the middle card here. How many cards do we have to put in B? Well, that’s 5 minus the card in position A minus the card that’ll be put in position C. That leaves us with 3 card choices. That’s our second number, 3.</p>

<p>(C), okay then. Let’s count together how many choices we have for this position. A was taken right? B was taken right? How many cards do you have left? That’s three cards. Any one of these three cards can be put in C, especially the one that we said could only be put in the middle. Okay then, that’s our third number, 3.</p>

<p>(D), it’s getting a bit repetitive from here isn’t it? Anyway, we’ve used up three of our cards on A, B, and C. We have two choices for position D. That leaves with the number 2.</p>

<p>(E), we’ve got only one card left to put here, so that’s leaves us at 1.</p>

<p>And now, the last step, multiply all possibles card positions to get the number of arrangements. OK, 4x3x3x2x1 = 72. And there we have our answer! :)</p>

5

<p>I freaking love you all. All of you. </p>

<p>Thanks so much!</p>