Permutations and Combinations: An Easy Method
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Ron is a Senior Instructor with The Princeton Review. Click here to read more articles from The Princeton Review and to learn more about The Princeton Review's GMAT services. |
If you’ve been studying Permutations and Combinations for the GMAT, you’ve probably had to memorize a bunch of ugly formulas involving factorials. Don’t worry – there’s an easier way.
Permutations vs. Combinations
First let’s go over the difference between Permutations and Combinations. These two terms are easy to mix up. Permutations how many different arrangements can be created from a group of people or items. Combinations ask how many different groups of people or items can be chosen from a larger group. Here are two examples:
In the Ugliest Dog competition, a blue ribbon will be awarded to the ugliest dog, a red ribbon to the second ugliest dog, and a yellow ribbon to the third ugliest dog. If there are 8 dogs in the competition, how many different ways can the ribbons be awarded?
This is a permutation question. Why? Because not only are we choosing three dogs, but the order in which they are arranged makes a difference. In other words, if the eight dogs are A,B,C,D,E,F,G, and H, and the ribbons go to dogs A,B, and C, it matters whether dog A is first, B is second, and C is third, or C is first, B is second, and A is third. In other words, A-B-C and C-B-A are considered two different outcomes. These are two distinct arrangements, or permutations. In fact, if you see the word arrangements in the question, you’re always dealing with a permutation.
Now consider this question:
Brandon wants to take 3 dogs with him on his morning walk. If Brandon owns 8 dogs, how many different groups of dogs can he choose for his walk?
This is a combination question because once Brandon chooses the three dogs, the order in which he arranges them doesn’t matter. In other words, if he calls dog A, then dog B, then dog C, or dog C, then dog B, then dog A, it doesn’t matter. It’s still the same group of dogs going on the walk. So if you see the word groups, you’re probably dealing with a combination question.
The Method
The method for doing these problems is simple. First, draw a series of underscores, one for each person or item you’re selecting. For example, in both of the above problems, we’re selecting 3 dogs, so we draw three slots:
___ ___ ___
If the question is a permutation, such as the Ugliest Dog example, you start by filling in the total number of items you have to choose from (in this case, 8,) in the first slot. Then count down as you fill each slot:
8 7 6
and multiply the numbers:
8 * 7 * 6 = 336 total permutations.
Combination questions, such as the one with Brandon and his dogs, are a bit more complicated. The only way to find the number of Combinations is by first finding the number of permutations, then dividing by the number of ways each group can be arranged. But don’t worry, there’s a simple method once again. Begin by doing exactly what you did in the permutation example. Draw three slots, start with the total number of choices, and count down:
8 7 6
But now on the bottom of the underscores, start with the number 1 and count up:
The number of groups of dogs that Brandon can take on his walk is:
That’s all there is to it. If it’s a permutation, just do the numbers on top of the slots; if it’s a combination, you have to do the numbers on top and the numbers on the bottom. There are just a few tips to remember:
- Permutations are when the order matters, combinations are when it doesn’t. If A-B-C is different from C-B-A, it’s a permutation; if A-B-C and C-B-A are the same, it’s a combination.
- The math is more complicated when the problem is a combination. This always confuses people: when the order doesn’t matter, the math is harder, because you have to do the numbers on the bottom. That might seem counterintuitive but that’s how it is.
- All of these problems assume that the same item cannot be used twice.
Here are two more examples:
The Gamma Zeta Beta fraternity must choose a committee of four members to plan its annual Children’s Hospital fund raiser and beer bash. If the fraternity has 10 members, how many different committees can be chosen?
This is a combination question because it only matters which 4 frat boys we choose. We are not arranging them in any particular order. So the solution is:
Remember, if the order doesn’t matter, you must do the numbers on the bottom! If you want to save time, reduce before you multiply: the 4 and the 2 on the bottom will cancel the 8 on top, and the 3 on the bottom reduces the 9 on top to a 3, so you have 10 * 3 * 7 = 210. The bottom can always be completely canceled out.
Now try this:
The Gamma Zeta Beta fraternity is electing a President, Vice President, Secretary, and Kegger Chair. If the fraternity has 10 members, in how many different ways can the officers be chosen?
Now we have a permutation, because we are not only choosing 4 frat boys, but also assigning them to particular jobs. So A-B-C-D is now different from D-C-B-A. This makes our math easier:
10 * 9 * 8 * 7 = 5040.
Leah on October 12th, 2009 at 8:55 am
This has been extremely helpful! I have been having such a hard time with these, Thanks!
Ritesh Bindal on October 12th, 2009 at 2:27 pm
That's really a very very easy way of explaining such topic which gives nightmares to many. I used to solve P&C problems, but with normal formulas like !8/!5 etc. But this is really helpful and easy to remember.
Thanks a lot!
Abhishek on October 12th, 2009 at 4:56 pm
Can anyone explain how to solve the one with repetition and specially says how many word you can form kid...
Ron Corcillo on October 13th, 2009 at 1:58 pm
For those you have to use a formula with factorials. Let's say you want to know how many 9-letter passwords you can form using 2 A's, 3 B's and 4 C's. The formula is 9!/(2!3!4!). The top of the fraction is the factorial of the total number of spots you're filling; the bottom is the factorials of how many times each element occurs (in this case, 2! for the two A's, 3! for the three B's, and 4! for the four C's.) The answer would be 1260. These don't show up that often but they are fairly easy if you know the formula.
Charlie on October 13th, 2009 at 12:24 pm
This is a fantastic article, I'm stunned at how simple this made combinations & permutations, the former of which I remember with a horrific formula:
(N k) = N! / (N-k)!(k)!
No more mwa ha ha, thank you!
Charlie on October 13th, 2009 at 12:37 pm
The question I have now is...what if the items can be used again?
For permutations, I think you just wouldn't subtract one. Suppose you had five colors of jewels, and you wanted to know how many ways you could set them in three positions on a ring, assuming you have unlimited of each color.
5 x 5 x 5 = 125
However, I'm unsure how to modify for combinations. How many different color combinations could you have?
125 / 6 doesn't equal an integer, so I'm assuming there needs to be a modification to the formula...
Ron Corcillo on October 13th, 2009 at 2:03 pm
You are right about how to do this for permutations. For combinations it's very complex. You have to look at 3 different scenarios: What if all three are the same color, what if two are the same and one is different, and what if all 3 are different. If all three are the same, there are 5 combinations (one for each color.) If two are the aame and one is different, there are 5 possible colors for the two that are the same, followed by 4 possible colors for the one that's different, which gives you 5x4 = 20 combinations. If all 3 are different, you would use the normal combination method, which is (5x4x3)/(3x2x1) = 10 combinations. You add these three different scenarios together, which gives a total of 5+20+10 = 35 combinations in all. It's very unlikely that you'll get one this ugly on the GMAT.
milind on October 13th, 2009 at 12:41 pm
You will get similar explanations in many places. This is the most simple part of P&C. I want to see the repetition part of it. That is where things start getting messy.
Ron Corcillo on October 13th, 2009 at 2:07 pm
Feel free to ask about more complicated problems, but keep this in mind: Perms & Combs are not that big of a topic on the actual GMAT. It is most likely that you will only see 1 or 2 of these problems and they will be relatively simple ones. A lot of people spend a lot of time worrying about these problems and then don't get any tough ones on the exam. If you are concerned about hard problems, better to focus on roots and exponents, number properties, and coordinate plane/slope of a line problems. These topics are much more common the real test, and they get just as hard.
Nasir A Khan on October 13th, 2009 at 6:26 pm
Appreciated !!! I had the problem of differentiating the permutation & Combination question. The easy way to remember in the hectic time.
Thank you for sharing.
Dilshod on October 15th, 2009 at 8:01 am
The article is very helpful indeed.
So far Ive been using those horrible formulas,but now it seems much easier to handle.
Thank you,
Good job,
Keep comming.
Sharat on October 22nd, 2009 at 7:46 am
Wonderful!! I have been using those mean factorial formaulas to solve such question but just knowing these tecgniques has helped me tremendously. Thanks!!
rajib on October 31st, 2009 at 11:25 pm
if there are 3 posts and 5 voters, how many ways the voter can cast their votes?
actually I wanted to know how to find out the n and r in the formula n^r. here what will be the permutation, 5^3 or 3^5?
SIR on November 1st, 2009 at 10:14 pm
Awesome!Always I have quit from my P&C lessons. For the first time I am ready to kick the ass of P&C. Thanks a lot bro RON!
Shilpi on November 21st, 2009 at 12:28 am
Oh My God!!! What an explantion! These topics were giving me nightmares. And the more I studied them, more confused I got. Do you have such lessons for roots and exponents, number properties, and coordinate plane/slope as well?
Thanks for your time.