PR Math Question

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<p>This is from Princeton Review, can anyone help me understand this concept? I see why I got it wrong but I still don't feel completely solid on all these combination/permutation questions.</p>

<p>The Tyler Jackson Dance Company plans to perform a piece that requires two dances. If there are 7 dancers in the company, how many different pairs of dancers could perform this piece?</p>

<p>I put 42 because I just thought 7 times 6, but the answers 21 because I guess you have to divide by two since they have to be different pairs since order doesn't matter.</p>

<p>Is there like a rule for this or something? Thanks in advance.</p>

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<p>Yes you are right, there is a formula for these type of problems. Here it is: (n!)/(n-k)!(k)! Where n is the total number of things you want to put in groups and k is the number of things per group. So for your example the formula would be used as (7!)/(7-2)!(2)! Which does in fact equal 21. Hope this helps!</p>

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<p>Perfect! Thanks man!</p>

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<p>

</p>

<p>Sometimes order matters, and sometimes it doesn’t. </p>

<p>If Cochran wins Survivor and Dawn comes in second, that’s a different result from Dawn winning and Cochran coming in second. Order matters. When order matters, you’re talking about permutations, and the formula for that is:</p>

<p>nPr = n!/(n-r)!, </p>

<p>where n is the number of people you’re choosing from, and r is the number of people that you’ll be choosing. That’s the formula you applied when you multiplied 7*6 and got 42.</p>

<p>If you’re just putting them into groups, however, then (back to Survivor) Cochran and Dawn is the same group as Dawn and Cochran. Order doesn’t matter; they’re the same group, no matter whom you name first. When order doesn’t matter, you’re talking about combinations, and you have to avoid double-counting. (Cochran & Dawn is the same as Dawn & Cochran, but both of those are part of your 42 permutations.) In order to eliminate double-counting, there’s another factor in the denominator of this formula: r!, because there are r! ways to arrange those r people you’ve chosen out of the larger group of n. So when order doesn’t matter, you need to use the combinations formula: </p>

<p>nCr = n!/[(n-r)!*r!].</p>

<p>Apologies if you were already fine with Remi’s answer. I’m just a wordy guy.</p>

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<p>I had a bit of an advantage: I’m a math teacher.</p>

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<p>You can do this computation very quickly with your TI-84 (or equivalent) calculator. Just use nCr in the MATH menu under PRB.</p>

<p>Type 7, then press nCr, then type 2, then ENTER. The output will be 21.</p>

<p>No need to memorize any formulas!</p>

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<p>Oh, yeah, I forgot that part. </p>

<p>When I’m teaching counting, I try not to mention that part to my class.</p>

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<p>In this case, I prefer to avoid the calculator AND the formulas!</p>

<p>I would rather tie it all back to the counting principle. So for example, if you are choosing president, secretary and treasurer from 10 people…</p>

<p>You have 10 choices for president, 9 for secretary and 8 for treasurer. So that thing we call 10P3 is just 10x9x8.</p>

<p>10!/(10-3)! = 10!/7! = 10x9x8. The formula is just a concise shorthand for what you get from the counting principle. But what you gain in conciseness, I find you often lose in understanding.</p>

<p>Then, if instead of officers, we just want a 3 person committee – so now order does not matter – then our first answer (10x9x8) overstates the answer by a factor that is equal to the number of ways to arrrange 3 things…so we have to divide (10x9x8) by(3x2x1).</p>

<p>So the thing we call 10C3 is also concisely written as (10!/(10-3)!)/3! or 10!/(7!x3!) but its really just the counting principle divided by what you can think of as a “redundancy correction factor”. </p>

<p>Certainly the formulas work, but once again, as you go to the formula, you make it less likely that you understand or even remember why any of this works.</p>

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<p>See its very easy ! I know people have already solved your problem but thats how I look at these questions :</p>

<p>7 dancers and 2 dances …</p>

<p>So which ever is the larger number becomes ‘n’ and the smaller number is ‘r’</p>

<p>so 7C2 that means 7x6 divided by 2</p>

<p>42/2 = 21 </p>

<p>I hope that helps. Sorry for bad grammar and punctuation. Its midnight in my country and I am sleepy.</p>

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<p>You are awesome. All of you. :)</p>

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<p>Here are a couple more to think about:</p>

<p>How many ways can you arrange the letters in the word THINK?</p>

<p>5x4x3x2x1</p>

<p>But how many ways if the word is HELLO?</p>

<p>5x4x3x2x1 is too big. It has redundancy – it treats the two L’s as if interchanging them makes a new order, which it does not. So it’s (5x4x3x2x1)/2 … that’s what I meant when I said to divide out the redundancy factor.</p>

<p>What about FREEZE?</p>

<p>Now, all of the arrangements that you get by re-arranging just the 3 E’s are redundant. And how many ways are there? 3x2x1 or 3! – so it’s 5!/3! — 3! is the redundancy factor.</p>

<p>So what about the letters in MISSISSSIPPI? I’ll let you do that one</p>

<p>None of this is on the SAT, but I’ve always liked these puzzles. And understanding the idea of dividing out the redundancy will help you to understand the difference between “permutations”, where the different orders count separately and “combinations” where you have to divide out the redundancy.</p>