<p>I need to review the rules for some counting problems.</p>
<li><p>How many ways can you seat x many people around a circular table?</p></li>
<li><p>I remember a question from Barron’s that asked something like - if there was a meeting at which each person shock every other person’s hand, and there was 28 handshakes how many people are there?</p></li>
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<p>CB’s counting problems seem to be much easier, but I suppose it can’t hurt to be prepared.</p>
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<li>I believe it is just x!</li>
<li>1+2+3+4+…+n would be the pattern. So just keep adding the next integer and you find that there are 7 people.</li>
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<p>^the answer for the handshake problem was 8… i did exactly what u did and got 7…</p>
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<li>isnt it a circular permutation problem, so isnt it (x-1)!?</li>
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<li> Are you sure? I thought it was less, because it was circular - it’s been awhile, so I could be totally mistaken :)</li>
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<p>what makes a circular table different from an octa-hexa-meta-pyramidal shaped table?</p>
<p>can sum1 confirm that question 1 is a circular permutaion problem? cuz if it is its npr/r or (n-1)!</p>
<p>The question about the handshake is a very simple one.
Since a hand shake consists of 2 people shaking hands, and you cannot repeat people, you would do : x C 2 = 28. Then you would make Pascals triangle and notice that this occurs when x = 8 .</p>
<p>^hahaha simple for you, maybe</p>
<p>Yes, I think rickymatsui is right about the first one - my response was to secret asian man. Isn’t there a way to do number two algebraically?</p>
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<li>for x p y, the answer is just</li>
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<h2>(x)! </h2>
<p>y! ( x-y)!</p>
<p>so in this case it would simply be</p>
<p>x!
----- = 28
2!(x-2)!</p>
<p>and since x!/(x-2)! is just x*(X-1) </p>
<p>(ex. if x was 4, 4!/2!= 4<em>3</em>2<em>1/2</em>1 cancels out to 4*3)</p>
<p>so x(x-1)=28*2=56
x^2-x-56=0
(x+7)(x-8)=0
x=-7, or 8</p>
<p>answer is 8 since -7 cannot exist in the real world </p>
<p>=]</p>
<p>Well, since x C y = x!/ (n-r)!r!, you can do x!/(x-2)!2! = 28, which then ends up with :
x(x-1)/2 = 28. x(x-1)=56. x^2-x-56=0 (x-8)(x+7) x= 8 or -7. You have to deduct the -7 since you can’t have -7 handshakes.</p>
<p>Oh, and for the first one, the answer would be x!.
Just do a couple of examples.
1 person = 1 way.
2 people = 2 ways.
3 people = 6 ways.
4 people = 24 ways.
etc. etc.
so its x!</p>
<p>For question 1
ok i looked it up on a website and i found:
[url=<a href=“3. Permutations (Ordered Arrangements)”>3. Permutations (Ordered Arrangements)]3</a>. Permutations (Ordered Arrangements)<a href=“scroll%20down”>/url</a>
Theorem 4 - Arranging Objects in a Circle</p>
<p>There are (n - 1)! ways to arrange n distinct objects in a circle.</p>
<p>so it should be (x-1)!</p>
<p>can anyone explain to me what the difference is between placing people in a circle vs a row?</p>
<p>^^^ yea in a circle u have to account for rotations
so u have to get rid of all the times u can rotate around a circile since its basically the same arangement</p>
<p>is every1 here taking math2 tm?</p>
<p>I am - At first I was kind of worried about Barron’s counting/prob/stats problems, because we have never covered those topics in school. However, CB’s problems were so easy I figured it wouldn’t really be necessary to completely understand these kind of problems 
Lets hope the test tomorrow is like their practice tests!</p>
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<li>(x-1)!</li>
<li>n(n-1)/2=28 , thus n=8</li>
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<p>Yeah… 10char.</p>