Boolean Problems and Combinatorics
INTRODUCTION
The GMAT contains a variety of logic problems dealing with set theory, number theory, binary categories, and combinatorics. While the technical terms for these categories seem a little daunting, most of the math involved is pretty basic; in fact, using math that is more compli- cated than necessary is a sure way to waste time and energy on the test.
BOOLEAN PROBLEMS ON THE GMAT
Boolean problems are those in which you are asked to determine whether a mathematical expression corresponds to one of two states. These questions could ask whether an expres- sion is
$ True or false
$ Even or odd
$ Positive or negative
$ Integer or noninteger
Clearly, there’s no way for any value to be bothof either of these; xcan’t be both even and odd, or positive and negative. That’s why the categories are “binary”; there are two distinct possibilities.
To answer this type of question efficiently, there are some rules that you absolutely, positively, must memorize if you haven’t already. Here they are:
Even vs. Odd
$ even× even=even $ even+even=even
$ odd× odd=odd $ odd+odd=even
$ even× odd=even $ even+odd=odd
Positive vs. Negative
$ positive× positive=positive $ positive+positive=positive
$ negative× negative=positive $ negative+negative=negative
$ positive× negative=negative $ positive+negative=It depends!
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When you add a positive and negative, the one with the larger absolute value (or distance from zero) wins. So if the positive number has the larger absolute value, the result will be positive, and vice versa.
Also, remember that subtracting a positive is the same as adding a negative, so if you see a subtraction sign, you can just change it to an addition sign, change the following number into its inverse, and follow the rules above.
Exponents
Exponents just continue all of the multiplication rules above. Therefore, assuming you’re dealing with integer exponents . . .
$ Anoddraised to an exponent continues to yield an oddresult, regardless of whether the expo- nent itself is even or odd: 32=9, 33=27, 34=81, 35=243.
$ Anevenraised to an exponent continues to yield an evenresult, regardless of whether the expo- nent itself is even or odd: 22=4, 23=8, 24=16, 25=32.
$ Any number raised to an evenexponent will be positive:−42=16, 32=9,−1– 22=1–
4.
Integers vs. Non-integers
integer× integer=integer
non-integer× non-integer=non-integer integer× non-integer=It depends!
When you multiply an integer and a non-integer, the result will be a non-integer unlessthe inte- ger is a multiple of the non-integer’s denominator. For example, 1/2×3=3⁄2(a non-integer), but 1/2×2= 1 and 1/2×4=2 (both are integers because 2 and 4 are multiples of the denominator 2). If that is confus- ing, remember that you can often use real numbers instead of rules to work these problems on the GMAT. And speaking of real numbers . . .
Other Tools for Working Problems Efficiently
Sometimes your knowledge of binary categories is enough to get you through a problem; other times, it’s a good way to ballpark out a few answer choices. When problems get more complex, though, it’s good to have a concrete way to deal with all of the abstract stuffthey like to throw at you. That’s why we recommend using real numbers when you can. Let’s try a couple of examples using the rules you memorized above (because you did, right?) and plugging in real numbers when you need to.
If x,y, and zare integers, then x2y2z will be negative whenever A. xis negative
B. yis positive C. zis negative D. xyis negative
If you’re totally on top of your Positive vs. Negative rules, including exponents, you may immedi- ately recognize that anything raised to an even exponent such as 2 will be positive, so both x2and y2will be positive . . . which means your result is positive so far. Only making znegative will give the negative result the problem is looking for, so the answer must be choice C.
However, if any one of those rules escapes your mind in this very stressful testing situation, what do you do? Well, you can use real numbers based on what the answer choices tell you.
• Choice A. If x=–2,x2is positive. You know you need a negativeresult in order for this answer choice to be correct, but you also know that if you can play by the rules in the problem and still get a positiveresult, you can eliminate this answer choice. So let’s try to get a positiveresult, substituting 2 for yand 3 for z(both positive numbers). Your final result is 4×4×3, and if you remember a few Positive vs. Negative rules, you don’t even have to multiply any further to know that the result is positive. Eliminate choice A.
• Choice B. Now you can use essentially the same numbers, this time making ynegative. How aboutx=2,y=–2, and z=3? Again, you get 4×4×3, a positive result. Eliminate choice B.
• Choice C. This time use positives for xandy, say 2 and 3. zmust be negative, so you can use –2.
Now your expression becomes 4×9×–2. You were able to get a negative result even using as many positive numbers as you could, so let’s keep that one.
• Choice D. Either xoryhas to be negative this time, because xyis negative. Well, you’ve already seen two examples in which either xorywas negative — answer choices A and B — and you were still able to get a positive result. So you can eliminate choice D.
• Choice E. Well, you wereable to tell what would cause you to get a negative result, in answer choice C. This answer is here only to confuse you. Eliminate choice E.
That means choice C is right again, no matter whether you use the rules alone or in combination with using real numbers. Either way, though, you can see that knowing the rules allows you to make some shortcuts. Let’s try another one.
xandyare positive integers. If xy+xis odd, then which of the following must be even?
A. x B. y C. x+y D. xy–x E. x2–y
Chances are that knowledge of the rules alone can’t take you all the way to the correct answer on this one, so let’s try some substitution. This time, let’s use a slightly different method for practice. We’ll pick real numbers to start, say x=3 and y=2, so that 3(2) +3=9, which is an odd result like the problem tells you it needs to be. Now you’re looking for answer choices that must be even.
• Choice A.x=3, which is not even, so you can eliminate choice A.
• Choice B.y=2, which is even, so you can keep choice B.
• Choice C. 3 +2=5, which is not even, so you can eliminate choice C.
• Choice D. 3(2) – 3 =3, which is not even, so you can eliminate choice D.
• Choice E. 32– 2 =7, which is not even, so you can eliminate choice E.
Luckily, you found only one answer that works. Choice B. But think back to Chapter 4; if more than one answer choice had been even, you would have just quickly substituted a different xandy (still ones that made xy+xodd) and kept narrowing down the answer choices that remained.
COMBINATORICS PROBLEMS: TWO APPROACHES
Combinatorics is the body of math that deals with combinations and permutations. Questions of this type are relatively rare on the GMAT, but you may see one, and if you are shooting for a high score, you will want to answer it correctly. Combinatorics problems tend to be conceptually difficult and time- consuming, so you need to approach them with a strategy if you don’t want them to eat up all of your time.
There are two approaches you can take to a combinatorics problem. The first could be called
“reasoned brute force”: understand what is being asked, reason through the problem, and then write down every combination or permutation asked for. On the GMAT these questions tend to involve fewer than 30 or so permutations, so it may be possible for you work out a problem like this in little more than 2 minutes if you dive right in.
The second approach is to solve the problems using combinatorics formulas. These formulas can be mathematically difficult and hard to apply, but they can give you the answer to a particularly difficult question far faster than brute force. We’ll discuss the brute force method first. Let’s consider a question:
In a family with 3 children, the parents have agreed to bring the children to the pet store and allow each child to choose a pet. This pet store sells only dogs, cats, and monkeys. If each child chooses exactly one animal, and if more than one child can choose the same kind of animal, how many different arrangements of animals could the family leave with?
A. 6 B. 8 C. 9 D. 10 E. 12
The first step with a problem like this is to figure out what you’re looking for. You’re looking for combinations of three variables—let’s call them D, C, and M—in groupings of three, in which a variable can be repeated more than once and in which order does not matter (i.e., DDC is the same as DCD). Let’s look at all the combinations the children could choose involving dogs:
DDD DDC DDM DCC DMM
So there is 1 possibility involving 3 dogs, 2 possibilities involving 2 dogs, and 3 possibilities involving 1 dog. Since all possibilities involving dogs are exhausted, it should be easier to pick out all the remaining combinations involving the remaining two variables:
CCC CCM CMM MMM
Can you think of any other possibilities? Let’s hope not, because there are no others. The answer is 10, which is choice D. Note that this system would have worked just as well if you’d started with monkeys or cats. The trick is to start on a system that will inevitably exhaust all possibilities.
Brute force has its limitations, though. Say you encounter the following question:
A sock drawer contains seven socks, each one a different solid color. The colors of the socks are red, orange, yellow, green, blue, indigo, and violet. If the socks are laid next to each other, how many different arrangements of socks could be made?
A. 24 B. 240 C. 1,024 D. 2,520 E. 5,040
First, understand what is being asked. You’re looking for permutations of seven variables—let’s call them R, O, Y, G, B, I, and V—in groupings of seven, with no repeat variables in each grouping, and order in this case does matter (i.e., ROYGBIV is not the same as ROYGBVI). You could write out every possible permutation, starting like this:
ROYGBIV ROYGBVI ROYGVIB ROYGVBI ROYGIVB ROYGIBV ROYBGIV ROYBGVI ROYBVIG ROYBVGI ROYBIVG ROYBIGV
ROYVBIG … and so on
As you can see, with some problems, the brute force method could take a very long time.
On this particular problem, it might take you the whole test, as you might be able to guess by looking at the scope of the available answers. It is in problems like this one that combinatorics formulas come in handy.
Factorials
The key concept in combinatoric math is the factorial. The factorial of an integer nis the successive product of ntimes all the positive integers smaller than n, right down to 1. In practice, that means that
n! = n× (n−1)×(n−2)× (n−3)×. . . 1
Actually, the 1 is wholly unnecessary, but you get the point. The factorial function blows up rather quickly, so you’ll not be asked to calculate a factorial much larger than these:
0! =1 (it is simply defined that way) 1! =1
2! =2× 1=2 3!= 3 ×2×1=6 4! =4×3×2× 1=24 5! =5×4×3×2× 1=120 6! =6×5×4×3×2× 1=720 7! =7×6×5×4×3×2× 1=5,040
Before we move on to the formulas, let’s consider how the concept of factorials alone could be used to solve the sock problem above. Suppose the question asked:
In how many different ways could you arrange a single red sock in a sock drawer if there are exactly 7 spaces for socks?
The answer, clearly, is 7. Now, what if the question asked:
In how many different ways could you arrange a single red sock and a single orange sock in a sock drawer if there are exactly 7 spaces for socks?
The answer is slightly more complicated this time. There are 7 ways you could arrange the red sock. Once the red sock is put in the drawer, there are now only 6 remaining spots in which to put the orange sock, so the total possible number of arrangements is 7 ×6 = 42. Now what if the question asked:
In how many different ways could you arrange a single red sock, a single orange sock, and a single yellow sock in a sock drawer if there are exactly 7 spaces for socks?
This time, we have 7 ways to put in the red sock, 6 places left to put in the orange sock, and then 5 remaining places left to put in the yellow sock. So the total number of possible arrangements is
Hopefully you see where this is going. To find seven unique spaces for seven unique spots in which the order matters, we will have:
7× 6× 5× 4× 3× 2× 1=5,040 possible arrangements
The answer to the question, therefore, is E. The factorial argument we just walked through is essentially a longhand version of the formulas we are about to discuss. The point of this explanation is that, even if you don’t remember the specific formulas to use, you can combine the concept of the factorial with your own mathematic reasoning to work through most combination and permutation problems.
Combinations
When we refer to combinationsinstead of permutations, we are talking about groupings in which order does not matter, e.g., a red sock anda green sock instead of a red sock and thena green sock. In the following formula, Cstands for the total number of possible combinations, nstands for the total number of elements involved (e.g., 7 socks), and kis the subset of unordered elements taken from n (e.g., 3 socks out of the 7):
This formula may look imposing, so let’s apply it to a problem:
A sock drawer contains seven socks, each one a different solid color. The colors of the socks are red, orange, yellow, green, blue, indigo, and violet. If a person reaches into the drawer and pulls out two socks, how many different color combinations are possible in the selected pair of socks?
A. 12 B. 15 C. 21 D. 36 E. 42
You could write out every possibility and solve this problem that way, but it is almost certainly faster to use the formula. In this case, n= 7 and k= 2:
Note that it was not necessary to calculate 7! out to 5,040. By canceling out common elements in the numerator and denominator, we were able to produce an equation that was quite simple to calculate. Most combination problems you see on the GMAT will work out smoothly like this; the test writers figure that if you can use this formula correctly, they don’t need to test your multiplica- tion skills.
nCk n
k n k
= −! = − = = × × × ×
!( )!
!
!( )!
!
!( )!
7 2 7 2
7 2 5
7 6 5 4 33 2
2 5 4 3 2
7 6 2
42 2 21
× × × ×× = × = =
nCk n
k n k
= −!
!( )!
Just for a visual verification, let’s consider how we would have worked this problem by brute force:
RO
RY OY
RG OG YG
RB OB YB GB
RI OI YI GI BI
RV OV YV GV BV IV
Add up the combinations: 6 +5+4+3+2+1 = 21. Clearly, either method works for a problem of this scope. One interesting thing to note with combinations is that picking 2 and leaving 5 behind is mathematically the same thing as picking 5 and leaving 2 behind. The formula for the latter situation would look like:
Now let’s move on to a slightly more complicated situation, in which the order of the elements matters.
Permutations
When order matters, the number of permutations is almost always larger than the equivalent number of unordered combinations. The formula for permutations, using the same variables of nfor the total number of elements involved and kas the ordered subset of n, is as follows:
This is described as “npickk,” and it is always a larger number than “nchoosek.” Note that, as opposed to the combinations formula, there is no k! on its own in the denominator. Let’s apply the formula to an ordered variant of our previous example problem:
A sock drawer contains seven socks, each one a different solid color. The colors of the socks are red, orange, yellow, green, blue, indigo, and violet. A person reaches into the drawer, pulls out two socks, and puts a single sock on each foot. If each sock can fit either the left or the right foot, and if it matters whether a particular sock is on the left or the right foot, how many different sock fashion combinations could the person make from the seven socks?
A. 12 B. 15 C. 21 D. 36 E. 42
Applying the formula, with n= 7 and k= 2, yields the following:
7 7 7 6 5 4 3 2
= ! = ! = × × × × × = × =
n kP n
n k
= −!
( )!
nCk n
k n k
= −! = − = = × × × ×
!( )!
!
!( )!
!
!( )!
7 5 7 2
7 5 2
7 6 5 4 33 2
5 4 3 2 2
7 6 2
42 2 21
× × × ×× = × = =
Working this problem by brute force would require writing down twice as many permutations as in the previous example if you wrote down every permutation, but you wouldn’t necessarily need to.
You could recognize that for each of the 21 combinations possible for pairing RO, RY, and so on, there is the mirror image form if the person switches the socks from right to left. In other words, for every RO there is an OR, for every RY there is a YR, and so on. Therefore, there must be exactly twice as many permutations as there are combinations, so 21 ×2 = 42.
Finally, let’s consider how we would apply the permutation formula to a previous question:
A sock drawer contains seven socks, each one a different solid color. The colors of the socks are red, orange, yellow, green, blue, indigo, and violet. If the socks are laid next to each other, how many different arrangements of socks could be made?
A. 24 B. 240 C. 1,024 D. 2,520 E. 5,040
Applying the permutation formula, with n= 7 and k= 7 (since there is no subset), we get the following:
Clearly, knowing the formula allows you to sidestep the ugly process of writing out over 5,000 different permutations. If you don’t feel that you can memorize the formulas, don’t worry. As we stated earlier, combinatorics problems are rare on the GMAT.
PRACTICE PROBLEMS
1. An art gallery owner is hanging paintings for a new show. Of the six paintings she has to choose from, she can only hang three on the main wall of the gallery. Assuming that she hangs as many as possible on that wall, in how many ways can she arrange the paintings?
A. 18 B. 30 C. 64 D. 120 E. 216
2. A composer’s guild is planning its spring concert, and ten pieces have been submitted for consideration. The director of the guild knows that they will only have time to present four of them. If the pieces can be played in any order, how many combinations of pieces are possible?
A. 40 B. 210
7 7
7 7 7
7 0
7 6 5 4 3 2 1
1 5 040
P = −! = = × × × × × × = ( )!
!
! ,
C. 1,090 D. 5,040 E. 10,000
3. If xandyare odd integers, which of the following must always be a non-integer?
A. xy B.
C.
D.
E. –xy
4. Dan has a membership at a local gym that also gives classes three nights a week. On any given class night, Dan has the option of taking yoga, weight training, or kickboxing classes. If Dan decides to go to either one or two classes per week, how many different combinations of classes are available?
A. 3 B. 6 C. 7 D. 9 E. 12
5. Terry is having lunch at a salad bar. There are two types of lettuce to choose from, as well as three types of tomatoes, and four types of olives. He must also choose whether or not to have one of the two types of soup on the side. If Terry has decided to have the salad and soup combo and he picks one type of lettuce, one type of tomato, and one type of olive for his salad, how many total options does he have for his lunch combo?
A. 9 B. 11 C. 24 D. 48 E. 54
6. If ris negative and sis positive, which of the following mustbe negative?
A.
B. s r s xy
2 y x x y