<p>I was practicing out of the blue book, and I found a question similar to this:</p>
<p>1 2 3 4 5</p>
<p>Suppose each number is written on a card and placed in a particular order. The number '2' cannot be on either end of the 5-card line. How many different variations are there?</p>
<p>Can someone explain how to do this?! Thanks</p>
<p>Combination problem.</p>
<p>Imagine there's 5 spaces. Each space can hold one number.</p>
<hr>
<p>In the first blank there can be a total of 4 different choices. (1, 3, 4, 5)
In the second blank considering you have chosen a number for the first, there can still be 4 choices.
In the third blank there can be 3 choices.
An so on...</p>
<p>So the math would be 4<em>4</em>3<em>2</em>1 = 96 unique possibilities.</p>
<p>That's what I was thinking, but the answer in the Blue Book says it's 72. Perhaps the book is wrong or maybe I'm missing a concept here.</p>
<p>By the way, this problem is labeled to be hard.</p>
<p>Imagine that there are no restrictions on the cards. There would be 5 P 5 number of ways to arrange them. This equals 120.</p>
<p>But the problem gives a restriction that a card can't appear on either end, so we have to subtract the arrangements that this occurs. If "2" is on the left end, there are 4 P 4 possible arrangements of the other cards. The same thing can be said if "2" was on the right end. </p>
<p>So the answer is 5P5 - 2(4P4) = 120 - 2(24) = 72.</p>
<p>What is wrong with gxing and my strategy?</p>
<p>Also, do you think you could help me figure out how to answer two other problems in the blue book (they have a diagram associated with them, so I don't know if I could type it up)? Thanks</p>
<p>1 possibility: 2 is in the second spot
then for the first position we chan select one number out of 1,3,4,5 which is one out of four
the second is full
for the third one we can select one number out of whats left which is three numbers
for the forth one we cen select one out of two
and for the last one we have only one left</p>
<p>so in this case we have 4x3x2x1</p>
<p>2 possibility: 2 is in the third spot
again 4x3x2x1</p>
<p>3 possiblity: 2 is in the fourth spot
4x3x2x1</p>
<p>so the answer should be 3 x 4x3x2x1 = 72</p>
<p>martinibluex, thanks for the explanation. I was never really good at combinations/permutations, so I'm not surprised if I get them wrong :( I understand how to get the answer now, but I'm afraid I won't get it if another variation of a combination or permutation problem pops up, especially in the open-end section.</p>
<p>If anyone has the blue book out and ready, here are two problems that I need help on!!</p>
<p>pg. 408 #5
pg. 427 #15</p>
<p>408 #5</p>
<p>when you spin the first time you can get 6 diferent numbers
when you spin the second time you can get 6 different numbers
so you have 6x6 = 36 - which is the number of all possible fractions you can get</p>
<p>the NUMBER OF THE "FAVORABLE" CASES is
when a=0 there's no such b
when a=2 then b can be only 1
when a=3 then b can be 2 or 1
a=4 > b=3,2,1
a=5 > b=4,3,2,1
a=6 > b=5,4,3,2,1</p>
<p>so favorable cases: 0+1+2+3+4+5=15</p>
<p>answer 15/36 I think</p>
<p>the two graphs are intersecting so you set the two Ys to be equal in order to find the Xs of the intersection points</p>
<p>so x^2=y=a-x^2
2x^2=a
x=+/-sqarert(a/2)</p>
<p>so the distance is the distance between sqarert(a/2) and -sqarert(a/2)
which is 2.squarert(a/2) (check out your x axis)
6=2.sqarert(a/2)</p>
<p>3=sqarert(a/2)
9=a/2
a=18</p>
<p>now you tell me how to become a faster reader :P</p>
<p>also another secret: even if you are not taking the SAT II math 2
go to Borders and just read the sections on Permutations and Combinations + Probability.
after that everything should look ok on SAT I math
good luck.</p>
<p>Wow. lol. I can't believe I did that problem wrong...</p>
<p>Another way of looking at the original problem is to put the cards in the slots in the following order: slot 1,5,2,3,4.</p>
<p>Slot 1 can use one of four legal cards; slot 5 can use one of three legal cards; and slots 2,3,4 can use one of (three)(two)(one) legal cards. So the total #variations is (4)(3)(3)(2)(1) = 72 .</p>