Strategies, Sample Questions, and Random Ramblings.
July 19, 2015
Permutations and combinations are two areas that most students do not want any part of, but as you will see it is really about developing a true understanding of what the formulas represent instead of memorizing the formulas. In all reality, in this section more than any other, I will encourage you to throw out the formulas used to calculate the solutions. First, let’s define what the difference between a combination and a permutation is exactly.
Combinations and permutations are both ways to find the number of arrangements in a group. The difference between the two is order. If order matters it is a permutation and if it does not it is a combination. Honestly, I hate that definition because it is difficult to understand. I prefer to think of it as a list and a group. A permutation is a list and a combination is a group. Basically, if you switch positions of two things in a list it is different, but if you switch two things within a group it is still the same. Thus, all else being equal, combinations will have fewer arrangements than permutations.
I will filter in the formulas along the way, but focus on your understanding of how the formulas connect to the way we set up the problem and you will have a great understanding of how to tackle these problems. Unlike all other texts, this book will start combinations problems and permutations problems in the same way.
Let’s start with a basic example of how many ways can you arrange the letters ABC?
If we were to list them all out they would look like this:
ABC ACB BAC BCA CAB CBA
Clearly, I do not want you to list all of these out each time, because it will get much larger. Instead, start every combination and permutation problem by creating a ‘space’ for each option. In this case, there are 3 letters we are picking so there are 3 spaces.
. . .
Then, simply place the number of options you have for each place. In this case you have 3 different options for the first space, then 2 for the second because you have already chosen one, and then just one for the last space. Then, take the product of all of these numbers.
3 . x 2 . x 1 . = 6
You will notice that this is simply 3!. It is true that most permutations come in the form of n! where n is the number of items to be arranged. This is the backbone to all of the problems that we will see going forward.
The next example is when we have less spaces than there are people or things to go in the spaces. Say that you have a race of 8 people, how many ways can gold, silver, and bronze metals be handed out?
8 . x 7 . x 6 . = 336
This is virtually identical to the scenario that we just did. Simply place the maximum number of items that can go into any space and then multiply. Some books will describe this formula as follows:
N is the same as it was before (the number of things to be arranged) and k is the number of spots to arrange them. This formula, however, is excessive work as by focusing on the spots as we already have, we eliminated the need for dividing (in all reality we already performed the operation).
So far we have only dealt with instances that are permutations. The next instance would be to select a group of 3 people from 5. Again, we start the same way of putting the maximum number of options in each available spot. The only difference is if you are dealing with a group, which we are in this case, you need to divide the product by the number of spaces you filled factorial - in this case three factorial.
The above scenario is your basic GMAT combination problem. The formula is as follows with n and k used to designate the same things as it did before.
The next next challenge you will face is when there are certain conditions or multiple combinations. Your key words to look for are “and” and “or”. If you see “or” you will have to add the results of separate combinations or permutations.
If Joe runs a race with 5 competitors, how many ways can the gold, silver and bronze medals be passed out if he finishes in first or second?
For questions like this you will want to break up the scenarios because there is conditionality. Basically, where Joe finishes has an impact on scenarios. For these questions deal with your constraints first, in this case Joe. Place him in each spot.
After you place Joe in each of his spots then fill in the remaining spaces with the remaining available options. Since these are two different scenarios, you will add the results for the total options - 24 in this case.
Another type of problem is a combined combination problem. In this case, “and” will signify that you have to multiply your results.
If a team of 2 men and 2 women is to be selected from 5 men and 4 women, how many different teams are possible.
10 x 6 = 60 possible ways.
It is important in scenarios like this to isolate the groups and divide out the repeats for groups individually before you multiply the results together.
Greg R., client, New York City
Emil C., client, Singapore
Chris S, client, New York City