## GMAT Tutorial: Combinatorics and Probability

### Combinatorics

Combinatorics is a topic that has enjoyed a fair amount of attention on the GMAT recently. Combinatorics is probably best thought of as a form of counting - counting the total number of possibilities for a given scenario.

In this tutorial, we will take a close-up look at combinatorics. We will start by examining the basics of combinatorics, first with a look at manual counting, followed by a study of a more formulaic method. Then we will take a look at counting when restrictions are placed on the scenarios. Finally, we will see how combinatorics applies to the field of probability.

Probability deals with the likelihood that some favorable occurrence(s) will happen. When it is necessary to count the number of favorable or total occurrences or both, combinatorics must be employed.

Let's begin our discussion with a look at a sample combinatorics question:

Three city council members, Mr. Gilligan, Ms. Appleton, and Ms. Murphy, are due to conduct a special committee hearing on secondary education. If three seats are designated for the council members, how many different possible seating arrangements are there?

This question is asking us to count the number of ways that three people can be seated in three different positions. Let's call the three council members G, A, and M.

One possible seating arrangement is GAM, where G sits in the 1st position, A in the 2nd and M in the 3rd. If we manually count the rest of the seating arrangements, we will see that there are a total of 6 possible seating arrangements.

• GAM
• GMA
• AMG
• AGM
• MAG
• MGA

Listing all of the possible arrangements is known as the process of "manual counting." An important point to notice about manual counting is that it is crucial to maintain a systemic method.

Always start by fixing a specific candidate in one position (usually the first). Then exhaust all of the possibilities that exist for that candidate in that position (in this case G followed by A then M or G followed by M then A). Then switch the candidate in the first position and repeat the process. Continue until you have finished with each candidate. In this way you will be sure that you have covered all possibilities.

Notice something important about this question: the different possible scenarios were generated SOLELY by a difference in the seating position of the council members. In general, in combinatorics questions one should always consider whether the question is just concerned about counting who will be in each group (i.e. which members) or whether it is also concerned about the ORDER of the members in each group.

In this question it was given that all three council members would attend the meeting (i.e. all scenarios have the same members); the only question that was left open was who would sit where. In the world of statistics, a combinatorics question that counts both which members and the order they appear is known as a permutation.

Typical scenarios that require order to be considered include: physical positions (seating, lining up items, etc), positions of function (president/vice-president, center/forward/guard, etc), and positions of reward (medals, ribbons, etc).

Let's take a look at another type of combinatorics questions:

Out of a group of 5 students (Ann, Bill, Charles, Diane and Ethan), 3 will be selected to represent the school on its debate team. How many different possible debate teams are there?

There is a fundamental difference between this question and the last in terms of what we are being asked to count. This question asks for the total number of teams. Teams are distinguished from one another solely on the basis of the members of the teams.

There is no significance to the order of the team members (no specific positions are indicated on the team). When we count the total number of teams, we will not consider the element of order:

• ABC
• ABD
• ABE
• ACD
• ACE
• BCD
• BCE
• BDE
• CDE

Just as there was a systematic method of counting the permutations in the first question, here too there is a method to the madness. Start with guaranteeing that the first member, A, is on the team and consider all such possible teams. Repeat the same by finding all teams where B is a guaranteed member and then again for C. You don't need to repeat this for D or E since there are no D or E teams that don't contain at least 1 of the first three members.

In this way you will be sure not to miss any possibility. Again, notice here that the 6 permutations for the same three member team (i.e. ABC, ACB, etc.) are NOT COUNTED. Order is not an issue here. The statistical term for this type of a combinatorics question when order doesn't matter is a combination.

The take-away from this chapter's discussion on the basics of combinatorics is twofold: (1) Manual counting is an important skill that you should always know to systematically employ if the need arises and (2) combinatorics questions can require you to count both the members in the group and the position they are in (i.e. permutations) or just the members of the group (i.e. combinations). Join us in the next chapter as we take a look at a more formulaic, automated approach to combinatorics questions on the GMAT.

### A Formulatic Approach Continued

The typical way to complicate a combinatorics question is by adding restrictions to the potential groups/arrangements.

Let's take another look at our city council members, this time with a restriction:

Five city council members, Mr. Gilligan, Ms. Appleton, Ms. Murphy, Mr. Roberts, and Mr. Thompson, are due to conduct a special committee hearing on secondary education. If Mr. Thompson and Mr. Roberts must sit next to each other during the meeting, how many different possible seating arrangements are there for the five council members?

There are five members to be seated: G, A, M, R, T If there were no restrictions on the seating arrangements, according to The Line Method there would be five lines for the five positions and the answer would be: _5_ x _4_ x _3_ x _2_ x _1_ = 5!

But Mr. Thompson and Mr. Roberts must sit next to each other. The easiest way to deal with a restriction of adjacent positioning is to consider the members that must be next to each other as one unit.

In this case we'll denote that unit as the R-T unit. We can now count the number of members to be seated as four: G, A, M, R-T. Using the line method we would then have four lines: _4_ x _3_ x _2_ x _1_ = 4!

But 4! isn't the final answer. What we've come up with by using four lines is almost all of the seating arrangements, for example GAMR-T, GAR-TM, GR-TAM, etc? Notice that wherever R-T is in the line-up, the order is always R then T.

The restriction, however, does not say that R must be seated before T. For every R-T seating arrangement, there is another, different T-R seating arrangement (GAMR-T and GAMT-R). For this reason we must multiply the total number of arrangements of 4 members (4!) by the total possibilities of the R-T arrangements (2!), so the final answer is 4!2! It is important to note that the multiplier here is 2!, not 2.

Using another example, if three of the five members had to sit next to each other (i.e. G, A, M-R-T), the answer would be 3!3! The first 3! is for the ways to position the three members (G, A, M-R-T); the second 3! is for the 6 different possible arrangements of the M-R-T subgroup.

A slight modification of this restriction of adjacent positioning is one of non-adjacent position, e.g. if Mr. Roberts and Mr. Thompson could not sit next to each other.

The easiest way to deal with this is to find the total number of unrestricted seating arrangements (determined above as 5!) and to subtract from that the number of adjacent seating positions (determined above as 4!2!). This gives us the total number of non-adjacent seating positions: 5! - 4!2!.

### Other Types of Counting Restrictions

Last chapter we began discussing restrictions in combinatorics questions, focusing on the restriction of adjacent positioning. Adjacent positioning is a restriction that applies to a permutation, which as you may recall is a combinatorics question that depends both on the members selected and their order/position.

In this chapter we will look at another typical restriction made to permutations - that of repetition among members. We will also see how restrictions can be applied to a combination, the type of combinatorics question that does not consider the order/position of the selected members.

A very common restriction that can be made to a permutation question is that of repetition among members. The classic example of this type of restriction is the anagram:

How many different words can be formed using the anagram MGMAT?

The best way to deal with this type of restriction is to first pretend that there is no restriction, i.e. no repetition of the letter M. Using The Line Method to count the total number of ways to order 5 letters, we would have _5_ x _4_ x _3_ x _2_ x _1_ = 5!

This count is slightly off however. If we call the two M's M1 and M2, the 5! has counted two identical arrangements as separate arrangements. For example, it incorrectly distinguishes between M1M2GAT and M2M1GAT. This incorrect counting is true for every arrangement so we must divide the 5! by 2! to eliminate the element of order between the M's.

Using another example, the number of words that could be formed from the anagram AAABBCC is 7!/(3!2!2!). Here we divide the 7! ways to arrange 7 letters by 3! to eliminate the element of order among the A's, by 2! to eliminate the element of order among the B's, and by 2! again to eliminate the element of order among the C's.

The technical term for this type of question is a permutation with repetition. Conceptually, this type of question represents an interesting hybrid of a permutation (order matters) and a combination (order doesn't matter).

Among different letters, order matters; among the repeated letters, order doesn't matter. Recall that for a combination question we divided the result of the line method by the number of lines factorial because we wanted to eliminate the element of order among all members. Here we only want to eliminate the element of order among the repeated letters so we divide by the number of times the letter(s) is/are repeated, factorial.

The last type of restriction that we will look at is for combinations. Let's return to our group of students, this time with a restriction:

Out of a group of 5 students (Ann, Bill, Charles, Diane and Ethan), 3 will be selected to represent the school on its debate team. If Ann must be on the team, how many different possible debate teams are there?

The key to this question lies in understanding the ramifications of the statement "Ann must be on the team." We solved this problem (in an earlier edition of this discussion) without the restriction and came up with 5!/(2!3!) different 3-member teams. What happens when Ann must be on the team? Actually the counting becomes much easier. We can take Ann out of the picture if she must be on the team; there's nothing to count with regards to Ann.

To complete each possible team, we must choose two people from the remaining four to join Ann. To count the number of two person teams that can be formed from a group of four, we set up the line method with two lines: _4_ x _3_ = 4!/2! Recall that because we don't care about the order of these two members, we must divide this expression by 2! (for the two lines) to eliminate the element of order. The final answer is 4!/(2!2!).

Join us in the next chapter as we see how counting principles can aid us in solving complex probability questions.

### Combinatorics and Probability

In this tutorial, we have taken a close-up look at combinatorics, both to provide you with an understanding of the nature of "counting questions" and to provide you with a formulaic approach for tackling these questions. We conclude our discussion of combinatorics with an application to a related GMAT topic - probability.

Probability deals with the assessment of chance - the likelihood that a certain scenario or set of scenarios will occur. In its most basic form, probability is expressed as a fraction (i.e. part/whole), where the part represents the number of favorable scenarios and the whole represents the total number of scenarios.

Probability questions can be made more complex in two basic ways. The first is by making the scenarios incorporate more than one event. For example, to find the probability of tossing a coin twice and getting heads twice in a row, we are now looking at a scenario that is comprised of two events - two independent tosses. The probability that the first event will occur (i.e. the result is heads) is ?. The probability that the second event will occur is also ?. The probability that the complex scenario of heads-heads will occur is ? x ? = ?.

In probability lingo, we refer to this as an example of "AND." The complex scenario is comprised of heads on the first toss AND heads on the second toss, so we multiply the two probabilities to find the probability of the complex scenario.

The second way that probability questions can become complicated is when the favorable scenarios incorporate multiple events that are also different from one another. Let's go back to the coins and consider the probability of getting one head and one tail when tossing a coin twice. (These are two different events that comprise one "winning" scenario.) It is tempting to say that the probability should be the same as the probability of getting two heads: ? x ? = ?.

What we've failed to consider here, however, are the two different scenarios for one head and one tail: TH or HT. Each of these scenarios has a probability of ? x ? = ?. The probability of getting a tail first and then a head would be ?. The probability of getting a head first and then a tail would be ?. The total probability of getting one head and one tail (either order) is the sum of the probabilities for these two different scenarios ? + ? = ?.

Notice that this is different than the probability of getting two heads because, while there is only one "order" for getting two heads (HH), there are two orderings for getting one heads and one tails (TH and HT).

Since there were only two orderings in the above problem, we were able to count them up in our head. However, what if we were asked to find the probability that on five coin tosses we would get three heads and two tails. Here, our knowledge of combinatorics becomes essential, since we don't want to count up all the favorable scenarios manually.

The number of different scenarios is the number of different orderings of HHHTT which, using the skills we learned earlier in this series, is equal to 5!/(3!2!) or 10. The probability of any ONE of these 10 scenarios occurring is equal to ? x ? x ? x ? x ? or 1/32.

To find the total probability, simply add up the probabilities of each of the 10 scenarios. This is equal to 10 x 1/32 or 10/32, which simplifies to 5/16.

As you can see, complicated probability questions often require knowledge of counting principles. This concludes our series on combinatorics!