<h1>22</h1>
<p>Why A is wrong:
There are no different sides presented in the third paragraph. All examples build on a single argument.</p>
<p>OK, here’s another counting problem, with two variants:</p>
<p>For a certain play, you have a regular hexagon painted on the stage floor and 6 actors who are to be assigned standing places at each vertex. Allen, Amy, Bob, Beth, Carlos and Cathy must be arranged so that each person stands at a vertex opposite a person with the same first initial. In how many ways can you do this?</p>
<p>Then, after the game, the six actors sit down to lunch at a circular table with 6 chairs, once again arranged so that each is opposite a person with the same initial. How many arrangements are possible?</p>
<p>[The difference between these two variants is subtle and I admit that I’ve never seen this point tested on the SAT. But for fun…]</p>
<p>6 x 4 x 2 x 1 x 1 x 1 = 48 different arrangements for the actors on the stage Hexagon.</p>
<p>Am I right?</p>
<p>What’s the difference between the first and second question too btw?</p>
<p>I think pckeller forgot to add seatings are considered to be the same if they can be obtained again by rotating the table…</p>
<p>Is the answer 8 possibilities for the second one?</p>
<p>I’m eager to know the answers… and some explanations!</p>
<p>pckeller, you make the greatest math problems up… 
Who are you!?</p>
<p>Thank you kindly!</p>
<p>But you don’t need my answers or explanations – yours are right on the money.</p>
<p>I am a high school physics teacher. But I have also been teaching math SATs forever and I wrote “The New Math SAT Game Plan.” And, as regulars on CC already know, I really like combinatorics…</p>
<p>As for these problems, I agree that it is hard to make clear that the circular table has rotational symmetry. Actually, hexagons have it too. That’s why I put the problem on a stage – so that it would be clear that rotating the actors would lead to a new, distinct arrangement with different actors in the foreground.</p>
<p>^ My answer is right, or JefferyJung’s is? :)</p>
<p>Also, Jeffrey, I didn’t quite grasp the answer to question two. Mind elaborating?</p>
<p>
</p>
<p>Yes… Your questions are ingenious :). I love the circular arrangement twist!
Do you make your problems up at that moment, or are these problems from your book?</p>
<p>
</p>
<p>Sure.
It’s actually pretty simple.
You got 48 possibilities, right? Well divide that by 6 since there are 6 rotations of the seating for every seating…</p>
<p>Or…
Seat one person and set this person as our basis. (this person will be seated with another person across from him/her)
Then, you do the counting for the other 4 seats…</p>
<p>4<em>1</em>2*1 = 8 arrangements.</p>
<p>I don’t think my explanation was too clear… Sorry!</p>
<p>The first one is 48 and the second one is 8.</p>
<p>The circle twist is not my invention – it is part of the traditional bag of tricks in a combinatorics chapter. N! ways to arrange in a line, (N-1)! ways to arrange in a circle…and (N-1)!/2 ways to arrange keys on a key ring. I’ll let you think about where the factor of 1/2 comes from. </p>
<p>I’ve posted a bunch of questions. A couple have been from my book but the more recent ones I just made up. But I’ll probably put them in the next edition.</p>
<p>
</p>
<p>I don’t think I’m really getting this, but is it 1/2 because the key ring can be flipped upside down?</p>
<p>Is the key ring like circular + 1/2(since it can be flipped)?</p>
<p>Yes! </p>
<p>So the arrangement</p>
<p>AB
CD</p>
<p>is the same as </p>
<p>AC
BD</p>
<p>flipped.</p>
<p>Just like each circular arrangement is one of N “rotationally equivalent” arrangements that you get by rotating everybody by one seat, each key-ring rotation is one of 2 arrangements that are “flipping equivalent”.</p>
<p>So…</p>
<p>(n-1)! For a circle is actually the same as n!/n (there are n rotations), right?..</p>
<p>This is very confusing but amusing at once. Haha.
Thanks for this combinatorics lesson!</p>
<p>You have it right: N!/N = (N-1)!</p>
<p>But for the record – I’ve never seen the circle twist on the SAT and I don’t think it is their kind of thing. Counting problems are hard enough as is. And the keychain flip is even more over the top.</p>
<p>Lol I’m liking the activity!</p>
<p>Here’s two math questions: If a,b,c,f are four nonzero numbers, then all of the following proportions are equivalent EXCEPT:
A) a/f = b/c
B) f/c = b/a
C) c/a = f/b
D) a/c = b/f
E) af/bc = 1/1</p>
<p>Ok, so I got this right because I saw that three of the answers lined up as one followed by the one two away from it… so the one where they were next to was the odd one out. The last answer was too obvious for a #19. But without looking at the answer choices, it is impossible to know how to do it, so that was a little weird.</p>
<p>Then #20: For all numbers x and y, let the operation [<em>] be defined by x[</em>]y = xy-y. If a and b are positive integers, which of the following can be equal to zero?</p>
<p>I. a[<em>]b
II. (a+b)[</em>]b
III. a<a href=“a+b”>_</a></p>
<p>A) I
B) II
C) III
D) I, II
E) I, III</p>
<p>
</p>
<p>I’m not understanding your thought process…</p>
<p>You can cross multiply each choices
A) ac = fb
B) fa = cb
C) cb = af
D) af = cb
E) af= bc</p>
<p>A is not equivalent to the other choices.</p>
<p>
</p>
<p>Note that a and b are positive. </p>
<p>I. a[_]b
= ab - b; Factor out b
b(a-1) = 0
b = 0; a-1 = 0
b cannot be 0
a-1 can be 0.
I. can be 0</p>
<p>II. (a+b)[_]b
= [(a+b)(b)]-b; Factor out b
b(a+b-1) = 0
b = 0; a+b-1 = 0</p>
<p>b cannot be 0, and a = 1-b
1 minus positive integer b cannot be a positive integer.
II. cannot be zero.</p>
<p>III. a<a href=“a+b”>_</a>
= [a(a+b)]-(a+b); Factor out a+b
(a+b)(a-1) = 0
a+b = 0; a-1 = 0
a+b cannot be 0 since a and b are both positive integers.
a-1 can be 0.
III. can be 0.</p>
<p>That was a hard one. I don’t understand how you did this:
(a+b)[_]b
= [(a+b)(b)]-b; Factor out b</p>
<p>Edit: Got it now, nevermind!</p>
<p>New question: Page 711 #24. How does the text imply that the answer is B and not A? Could be either imo</p>
<p>
</p>
<p>[(a+b)(b)]-b; distribute b
= ab+b²-b; factor out b
= b(a+b-1)</p>
<p>You can do this, or think of a+b as another variable X (a+b = X)
Xb-b; factor out b
= b(X-1); X = a+b
= b(a+b-1)</p>
<p>
</p>
<p>How can it be A? It’s actually the opposite. The author suggests that people have the propensity to isolate themselves.</p>
<p>On a side note, I actually got this problem wrong. 
I chose E, but E is apparently too extreme.</p>
<p>Actually yeah now that I think about it, I see why A is wrong. I was thinking of the people around them being hindrances</p>