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GMAT Combinations: Consider The Grouping with the Slot Method

GMAT Combinations

The GMAT loves to present combination questions that force people to sit next to each other, or in certain seats, and then have the test taker figure out how many different ways people not forced to be glued to possibly a very uncomfortable chair can be seated around them. Because free will, clearly, is not an option for many movie-goers, committee members, and/or people seated in a kumbaya circle.

Take this question, for example:

A group of five students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?

  • 5 percent
  • 10 percent
  • 15 percent
  • 20 percent
  • 25 percent

See where we are going with this post?

Many test takers will immediately jump to a combination, or potentially a permutation formula (you know, because the order matters), and find themselves completely stuck on what inputs to use. Is it five? Or three? because Bob and Lisa don’t like middle seats?

The reality is that using the slot method–sketching out the seated relationship between these five people–will make it very apparent a). how much easier this problem is to tackle and b). help direct test takers more quickly to the right method of calculation rather than being mired down in trying to figure out if n!/(n-k!) works and what numbers should be used for n and k.

For this question with Bob and Lisa, and similar types of questions, try sketching out as follows:

Bob _____ _____ _____ Lisa

OR

Lisa _____ _____ _____ Bob

It looks like that there are only two different ways that Bob and Lisa can be seated, so then the focus becomes how the remaining three students can be seated. This is accomplished with a very simple factorial–3!–representing the different combinations of the remaining three students.

Alternatively, while a little more time-consuming, considering the combinations of those three movie-goers (who we can name Amy, Belinda, and Carl) is another route that helps us get to the right answer.

ABC

BCA

CAB

BAC

ACB

ACB

Producing six unique combinations. Multiple those six by two–the different seats that Bob and Lisa can be in–and we have 12 different arrangements of student movie-goers. So, 3! = 6 x 2 = 12.

But the issue is that the answer choices are all percentages. By re-evaluating the question, the test taker is being asking to figure out what the probability of this combination will turn out to be. This can be calculated by:

Number of total outcomes where Bob and Lisa only sit next to one person

Total number of possible outcomes

With five students, 5! = 120 is the total number of combinations without restrictions. Placing 12/120 = 10 percent making our answer (B)

When thinking about these types of questions–combinations with restrictions–consider the slot method. Your path to the right answer may be simpler than you think!


This article has been edited and republished with permissions from its original source, Clear Admit.

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