Permutation / Computation Guide

Test Preparation
1

<p>Hey guys, I need an overall guide of permutation / combination. I always get stumped on these kinds of problems, so is there an easy way to tackle these problems? Unfortunately, I don't have an example to use, but if someone does please feel free to post. Whats the formula to use and/or shortcut etc. Please go in depth because this is one of my weaker points. It would save a lot of time if you could show an easy way to approach them. Thanks</p>

<p>EDIT - actually heres an example:</p>

<p>1 2 3 4 5</p>

<p>Imagine each number represents a card, and the card number '2' can never be on either end, how many different assortments are there?</p>

2

<p>For your example: The number 2 can't be on the ends so: nPr(4,2) would be the number of ways that the remaining 4 numbers can be put on the ends. Then for the middle three, there are 3! or 6 ways to arrange the remaing 3 numbers. Thus, there are 6*nPr(4,2)= 72 ways. Look at all problems in a semi-similar way to what i did. just think, what is the problem asking and how can it be simplified and expressed mathematically?</p>

3

<p>Thanks for the quick response...but unfortunately I'm slow and don't really get that because I don't know what nPr means what so ever. If you could go into more detail I'd really appreciate it. thx</p>

4

<p>1337, how do I calculate nPr(4,2)? I remember there was a longer form of it, like n!/(n-r)! or something like that. Also, what do the parenthesis represent and how do I use that to calculate a numerical answer?</p>

<p>Sorry, I was never good at combinations/permutations :(</p>

5

<p>In order to be good at combinations/permutations, you must understand the logic behind them. When we learnt them in class, we started off drawing little boxes for each possible place. Then, when we understood what possibilities there were for each box, my teacher taught us nPr/nCr. To try doing those without understanding the basics is silly because you are much more likely to make mistakes then. So I'll try explaining how we first learnt them but I might not be too clear as I don't explain math very well. :-P</p>

<p>First, draw 5 little boxes from left to right. In the first box to the left, only 4 of the 5 cards can be placed there since the number 2 cannot be at the end. Next, skip over the 3 middle boxes and consider the last box. Since the number 2 cannot be there AND since you already placed one card in the box all the way to the left, there are only 3 possible cards that can go there. Now, go back to the second to left box. Since you already got rid of 2 cards (one at either end) AND since the number 2 CAN go there, there are 3 cards that can possibly be put there. As for the third box to the left, you only have 2 out of the 5 cards left and either can be put there. Finally, for the fourth box, you only have 1 card left! Now, simply multiple the possibilites and you get 4<em>3</em>2<em>1</em>3 which equals 72. Sounds a bit long but once you get the hang of it, it is really simply to do.</p>

<p>I do not recommend using the nPr and nCr formulas because you can get confused easily if you haven't practiced much with them. I hope this helped a bit. :-D</p>

6

<p>wisywigi, I know exactly what you're saying when you mean 'drawing boxes, etc', because I did it in 7th grade (kinda like a primative version of nPr/nCr). It's just really sad that I can't understand this right now.</p>

<p>Can you give me an example problem for me to try? I feel like combinations/permutations are the biggest things holding me back :(</p>

7

<p>Thanks for trying to explain, but I still don't get it. Maybe I'm overanalyzing or something, but I'm confusing myself. I understand how 2 can't be placed in either end so of the 5 cards..1 of 4 can be placed on the left and 1 of 3 can be placed on the right. Then you're left with '2' card and 2 others. Those can be placed anywhere in the middle 3 slots. I just don't understand how you're supposed to multiply / do the math work once you understand that. Nor do I understand the difference between permutation / combination and what nCr and nPr mean (the n / r means?). Thanks for the help.</p>

8

<p>Subtract the number of possible permutations containing 2 on either end from the total number of permutations.</p>

<p>So:</p>

<p>2 x x x x, There are 4 items, so unique permutations with 2 on the left side end is 4!, which is 24. There is an equivalent number of unique permutations for the 2 being on the right side. Therefore, total permutations with 2 on either side is 4! x 2 which is 48. </p>

<p>Total permutations is 5!, so 5! - 48 = 72.</p>

9

<p>THANKS - that explains is a lot easier...is this always the case? subtracting the restricted permutation from the total permutations? Also can someone explain how to use nPr / nCr. thanks</p>

10

<p>Yeah, usually in situations where you want to find the number of permutations possible with a certain limitation, it is good to find the ones noninclusive to the limitation and subtract it from the total.</p>

<p>Permutation is simple, it is the different arrangements of a set of elements, such as: 123, 321, 231, 213</p>

<p>So a permutation of 4 elements is 4 x 3 x 2 x 1 because you have 4 choices for the 1st item, 3 for the 2nd, 2 for the third, and 1 for the last. (Note there can be repeated elements in each arrangement because the 1st for one arrangement might be picked as the 2nd or 3rd or 4th in another arrangement.) The equation is 4!</p>

<p>Thus, nPr stands for n elements choose r elements for a Permutation. (P stands for Permutation.) And the formula would be n!</p>

<p>A combination is a set of unique elements with no consideration towards the sequence.</p>

<p>nCr means n items, Choose r for a Combination. (C stands for combination.) So a combination is n!/(n-r)! (r!)</p>

<p>If you don't understand why this formula is the way it is, think of it this way:</p>

<p>If you want to pick 3 items from 4, you will have 4 for the 1st choice, 3 for the 2nd, and 2 for the 3rd.</p>

<p>4!/4-3 is equal to this. It cancels out the factorization up to the first 3 or first r factors.</p>

<p>So it becomes: 4 x 3 x 2</p>

<p>However, this is still a permutation. This is because the first item can be picked as the 2nd item if it were not picked as the first. This goes for all elements modeled this way and thus it has repeats. A combination does not have repeats. So simply divide the permutation by the number of permutations 3 (or r) elements can have. This gives you the number of unique combinations by cancelling out all of the different permutations with the same elements. </p>

<p>So the nCr equation is n!/(n-r)!r!</p>