Chapter 1: Introduction to Permutations Print out chapter Permutation questions are about taking a group of objects and totaling how many ways we can arrange them in specific ways. Here is an example that we will explain later. In how many ways can a pet shop line up 3 cats and 3 dogs in 6 cages if the cats must be in the second, fourth, and sixth cages? I. The Basics: Three Steps to Permutation Clarity 1. Figure out how many places there are to fill. 2. Figure out how many objects potentially can go into each place. 3. Multiply for the answer. Example How many outcomes are there when two identical dice are rolled? Following the steps: 1. Figure out how many places there are to fill Because there are two dice, there are two places to fill: __ __ 2. Figure out how many objects potentially can go into each place Because each die has 6 different potential outcomes, we will fill the spaces accordingly: _6 _ _6 _ 3. Multiply for the answer _6 _ × _6 _ = 36 Example 2 In Country X, three digit area codes are to be given to each town. The first digit will be any number from 2-9, inclusive, the second digit can only be either 0 or 1, and the third digit can be any number from 0-9, inclusive. How many different area codes can be issued in Country X? Following the steps: 1. Figure out how many places there are to fill
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Chapter 1: Introduction to Permutations
Print out chapter
Permutation questions are about taking a group of objects and totaling how many ways we can arrange them in
specific ways. Here is an example that we will explain later.
In how many ways can a pet shop line up 3 cats and 3 dogs in 6 cages if the cats must be in the
second, fourth, and sixth cages?
I. The Basics: Three Steps to Permutation Clarity
1. Figure out how many places there are to fill.
2. Figure out how many objects potentially can go into each place.
3. Multiply for the answer.
Example
How many outcomes are there when two identical dice are rolled?
Following the steps:
1. Figure out how many places there are to fill
Because there are two dice, there are two places to fill:
__ __
2. Figure out how many objects potentially can go into each place
Because each die has 6 different potential outcomes, we will fill the spaces accordingly:
_6_ _6_
3. Multiply for the answer
_6_ × _6_ = 36
Example 2
In Country X, three digit area codes are to be given to each town. The first digit will be any number from 2-9,
inclusive, the second digit can only be either 0 or 1, and the third digit can be any number from 0-9, inclusive.
How many different area codes can be issued in Country X?
You eliminate all doubles (shown on the diagonal above) [(1,1), (2,2), (3,3), (4,4), (5,5),
(6,6)] because the game says that if get the same number, the results aren’t counted. So
the result is 30 different permutations. Just count all of the outcomes above to get 30.
II. Example of Combinations
Now let’s do the “combinations” versions of the above question.
We’ll try the game again, but instead of using a red die and a blue die,
this time both dice are red. How many different possibilities are there?
First, we can easily identify this as a combinations question because order or placement
isn’t an issue. In the first question a (5,2) is different from a (2,5) because they have
different colors. This time, they are both the same color, so you can’t tell a (2,5) and a (5,2) apart, so order doesn’t matter. Once order or position doesn’t matter, this becomes a combinations question and not a permutations question.
(Red,Red)
(Red,Red)
(Red,Red)
(Red,Red)
(Red,Red)
(Red,Red)
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (1,3) (3,3) (4,3) (5,3) (6,3)
(1,2) (1,2) (3,4) (4,4) (5,4) (6,4)
(1,3) (1,3) (3,5) (4,5) (5,5) (6,5)
(1,4) (1,4) (3,6) (4,6) (5,6) (6,6)
This chart shows all the possibilities (30) but now you can see that half of the outcomes
are redundant. For example, we have counted (4,3) and (3,4) (bolded above) when the
two red dice will look the same. One of them can't be counted to the total number of
combinations. You have to eliminate the double counted results in combinations
questions.
Therefore, we have to reduce the total number of permutations. All the possibilities that are overlaps/double counted in an earlier column we will strike through.
(Red,Red)
(Red,Red)
(Red,Red)
(Red,Red)
(Red,Red)
(Red,Red)
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (1,3) (3,3) (4,3) (5,3) (6,3)
(1,2) (1,2) (3,4) (4,4) (5,4) (6,4)
(1,3) (1,3) (3,5) (4,5) (5,5) (6,5)
(1,4) (1,4) (3,6) (4,6) (5,6) (6,6)
If we count the final results, we get only 15 total combinations (in bold red). Obviously
you can’t always rely on using charts like this if the number of possibilities is much larger,
so there must be a simpler way to solve these combinations problems.
1. Do the problem as if it was a permutations problem.
2. Divide the answer by (the number of spaces)! (factorial)
You can combine the above two steps into this simple combinations formula:
This formula is very similar to our permutation formula. In this problem, the n stands for
distinct objects to choose from, r stands for the spaces into which n objects can fit, and
the C stands for stating that this is a Combinations problem. The only difference between
is the addition of r on the bottom.
Let’s apply this to our dice problem. We have two dice, so there are two spaces.
Following the combinations formula:
Notice that there are only 15 combinations while there are 30 permutations. Having red and blue dice made 15 more possibilities than red and red dice.
800score Tip:
Combinations are always fewer than permutations.
The implication of the formula above is that combinations
are smaller than permutations of the same numbers
because we always start with permutations and then divide.
Use logic: combinations mean results that are the same are
double counted and therefore don't count, so they need to
be excluded. Combinations don't address order and
therefore produce fewer possibilities.
Chapter 2: Permutation or Combination?
Chapter 2: Permutation or Combination?
Print out chapter
800score Tip:
Half of the challenge in most combination/permutation
questions is correctly identifying if it is a combination or
permutation question. Once you know this, most become
easy.
1. Does order matter?
“Does it matter if the two items (or however many items you have) positions are changed?” If the answer is yes,
it is a permutations problem, if not, it is a combinations problem.
2. If the word “arrangements” is in the problem, it is a permutations problem.
The term “permutations” will rarely be in a GMAT problem.
3. If the word “combinations” is in the problem, it is a combinations problem.
Don't think it'll be that easy often!
Here are 7 examples each of permutations and combinations. Read through them and try to think out each one. You will soon see the difference between
3. Multiply the answers together for the final answer.
70 × 35 × 10 = 24,500
We can also use the combinations formula to solve for this:
nCr × nCr × nCr = 8C4 × 7C3 × 5C3
For this problem, we need to set up three separate combinations equations and then multiply them all together
to get the final answer. Our first equation represents the 8 meats Jonathan can put into the four spots on the
pizza. The second represents the 7 vegetables for the 3 spaces available, and the last equation represents the
5 cheeses that can go into the 3 available spaces on the pizza. Now let’s expand out the equation:
And now we multiply each result to get the final answer, and we get:
70 × 35 × 10 = 24,500
Example 11
A committee of 7 members will be chosen from 3 groups:
3 from Green Group, which has 6 people
3 from Red Group, which has 8 people
1 from Purple Group, which has 5 people
How many different committees can be created?
1. Do the problem as if it were a permutations problem.
Green Red Purple
_6_ × _5_ x _4_ _8_ × _7_ × _6_ _5_
2. Divide the answer by (the number of spaces)!
3. Multiply the answers together for the final answer.
20 × 56 × 5 = 5600
We can also use the combinations formula to solve for this:
nCr × nCr × nCr = 6C3 × 8C3 × 5C1
Once again, we need to set up three separate combinations equations and then multiply them all together to
get the final answer. Our first equation represents the 6 people we can choose for the 3 spots from the Green
group. The second represents the 8 people from the Red group available for the 3 spots on the committee, and
the last equation represents the 5 people from Purple that can go into the 1 space on the committee.
Now let’s expand out the equation:
And now we multiply each result to get the final answer, and we get:
20 × 56 × 5 = 5600
Variation 2: Pairings
The title says it all: these are pairing questions – so they must take place in twos.
Example 12
6 people in a room each shake hands with one another. If no one shakes hands with any other person more
than once, how many handshakes take place?
Let’s approach this with a discussion. How many spaces should there be? Many people want to put six. But
actually, there are only 2. Why? Because there are only 2 people involved in any handshake!
Solution
In this case, there are 6 people who could be the first person, and then five people to shake that person’s
hand. So following our approach for combinations:
1. Do the problem as if it was a permutations problem.
_6_ × _5_
2. Divide the answer by (the number of spaces)!
There are two spaces, so divide by 2!
6 × 5
= 15
2 × 1
You can also solve for this by using the combinations formula:
nCr = 6C2
Where n stands for the 6 people that we are choosing from to shake hands, and r stands for the 2 people who
are actually shaking hands.
Example 13
7 basketball teams with five players each are at a tournament. If each player shakes hands with every other
player NOT on his own team, how many handshakes take place?
Explanation
This one is MUCH trickier than the last one. But the logic remains the same. How many people take part in a
handshake? Two. So there must be two spaces. But in this case we have to use the logic of the problem to
answer it correctly. There are 35 people who could be in the first spot, but that person cannot shake hands
with anyone on his own team. So that person has only 30 people who’s hands he can shake! That will be
reflected in the combination. So let’s approach it using the method and this logic:
1. Do the problem as if it was a permutations problem.
_35_ × _30_
2. Divide the answer by (the number of spaces)!
There are two spaces, so divide by 2! And don’t forget to cancel!
GMAT Probability: Simple Probability
Chapter 1: Simple Probability
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Probability is often seen as the real evil of the GMAT. But the truth is that by mastering one simple fraction, you
can make probability approachable and even easy.
Let’s say that you have a single six-sided die. If you role it, what is the probability you would
roll a 5?
There are 6 sides to the die, and each one could come up. The 5 side is one of those sides, right? So there are 6 possible outcomes, and the five is one of them. Therefore, there is a 1 in 6 probability that the five will come up.
800Score Technique: The Bottom and the Top
If you want to make probability approachable, just think of it
as a fraction to solve Bottom to Top. The bottom number is
the total number of possibilities that could happen, and the
top number is the number of possible ways to achieve the
Let’s go right back to the first example in this chapter:
Let’s say you have a single six-sided die. If you role it, what is the probability you will roll a 1?
If you remember, the answer was 1/6.
Now let’s put some money on this. You’re in Vegas, and you’re going to win $500 if the die lands on 1.
When it does, you just won $500! Let’s call that a success. You had a 1/6 chance to achieve that success.
But you decide to let it ride! You’re convinced it can happen again. You go double or nothing, betting again
on 1. This time, the die comes up on something else! You’ve failed. And since 5 out of the 6 outcomes
would have caused that failure, the probability to fail is 5/6.
Now, let’s turn it around. What if someone were to offer you $500 if, when you roll one die, you rolled at
least a 2? Think about that for a second. What outcome would win you $500? If you’re just thinking of 2,
think again. In fact, if you rolled a 2, or a 3, or a 4, or a 5, or a 6, you would win! That’s pretty good. What
are the odds of that happening? You could figure out that there are 5 ways to win out of 6, so the answer
would be 5/6. The probability of success would be 5/6.
But, by using the phrase “at least,” the problem created more ways to succeed than to fail. That means
more work. On the GMAT, we want to avoid hard work. Try to figure out how you could fail, and then
reverse it? The only way NOT to win $500 is to roll a 1. And what’s the probability that you’d roll a 1? 1/6.
The probability of failure is 1/6, so the probability of success must be 5/6.
To simplify it, look at this formula:
P(success) + P(failure) = 1
or
P(success) = 1 – P(failure)
Example 15A jar contains 10 red marbles and 6 black marbles. If three marbles are pulled from the jar one after another without replacing them back into the jar, what is the probability there will be at least one red marble among them?
Answer: 27/28
Can you see why this would be tough? There are so many ways to pull out “at least one” red marble. The
red one can come first, second or third. Or, you could pull out two red marbles and one black one, in any
order, or you could pull out three red marbles. Figuring out the total number of ways to do that would take a
very long time.
But, think about the failure in this case. What is the one way to fail? You can fail by not pulling out any reds
Answer: Think Bottom to TopBottom: We have two dice being rolled, with 36 total outcomes possible.
Top: There are two outcomes that can happen: (3,4) and (4,3).
Probability: 2/36 = 1/18
Example 18What is the probability that rolling two identical dice together will result in the sum of the dice adding to 7?
Answer: Think Bottom to TopNote: We saw this question above in example 14. Try solving it a different way.Bottom: Total outcomes = 36
Top: How many of those outcomes yield a 7? Just list them:
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
There are 6 in total.
Probability: 6/36 = 1/6
Example 19Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, numbered 5 - 9. What is the probability that the sum of a card pulled randomly from Bowl X and a card pulled randomly from Bowl Y will equal 8?
Answer: Think Bottom to TopBottom: How many different pairs of cards can be pulled? There are 4 cards in Bowl X and 5 cards in
Bowl Y. Using permutations: 4 × 5 = 20. So there are 20 total outcomes possible.Top: Which pairs work for this question?
(1,7) (2,6) (3,5)
3 outcomes.Probability: 3/20
Example 20Bowl X has 4 cards in it, numbered 1-4. Bowl Y has 5 cards in it, numbered 5-9. What is the probability that the product of a card pulled randomly
from Bowl X and a card pulled randomly from Bowl Y will be even?
Answer: Think Bottom to TopBottom: 20 total outcomes
Top: Which pairs will multiply to an even number?
(1,6) (2,8) (4,6)
(1,8) (2,9) (4,7)
(2,5) (3,6) (4,8)
(2,6) (3,8) (4,8)
(2,7) (4,5)
14 total pairs.Probability: 14/20 = 7/10
Alternative Answer Think about success and failure. What is a success in this problem? An even outcome. How can two
numbers multiply together to make an even? There are lots of ways – all we need is an even number. What
if we reverse this one? Since there are only two outcomes (odd and even), it should be clear that there are
less odd pairs than even ones. Odd outcomes are failures. Let’s calculate the probability of failure, and
then subtract from 1 to find the success:
(1,5) (3,5)
(1,7) (3,7)
(1,9) (3,9)
Bottom: 20 total outcomes
Top: 6 odd outcomes
Probability of Odd Outcome (failure): 6/20 = 3/10
Probability of Even Outcome (success): 1 – 3/10 = 7/10
Example 21A fair coin is flipped three times. What is the probability that heads will come up only once?
Answer: Think Bottom to TopNote: We saw this question above in example 11.Bottom: How many different outcomes are possible? Each time the coin is tossed, there are two
outcomes, and we’re tossing it three times: 2 x 2 x 2 = 8Top: How many outcomes have only a single instance of heads? There are only 8 possibilities, as we
learned above, so there can’t be too many. Work it out:
H T TT H TT T H
3 total outcomesProbability: 3/8
Chapter 6: Extra Questions
Chapter 6: Extra Questions
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1. A jar has 10 marbles, either black or white. 2 marbles are randomly chosen from the jar. If q is the
probability that both will be black, is q > 1/3?
1) Less than 1/2 of the marbles in the jar are white.
2) The probability that 1 white marble and 1 black marble will be chosen together is 7/15.
A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though
NEITHER statement BY ITSELF is sufficient.
D) Either statement BY ITSELF is sufficient to answer the question.
E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that
further information would be needed to answer the question.
SolutionThis is a dependent probability problem. If you want to find the probability of choosing 2 black marbles, you will need to figure out the probability that the first marble will be black and that the second marble will be black. In this case, the question wants to know if that probability is larger than 1/3.
Statement 1 tells us that less than half the marbles are white, which
means that more than half the marbles are black. The best way to approach this is to systematically (but quickly) figure out what the probability of two black marbles is for each scenario. We can do it easily by drawing a chart:
Black Marbles White Marbles P(2 Black)
6 4 6/10 × 5/9 = 1/3
7 3 7/10 × 6/9 = 21/45
8 2 8/10 × 7/9 = 28/45
9 1 9/10 × 8/9 = 4/5
10 0 10/10 × 9/9 = 1
As you can see, when less than half the marbles are white, the probability of choosing 2 black marbles can
be higher or equal to 1/3, depending on how many black marbles there are. This is not sufficient. Statement 2 tells us that the probability of choosing one black marble and one white marble is 7/15.
This is a trap. Since the probability given is exact, it may seem that only one scenario of black marbles and
white marbles will work. If you work through all the scenarios, you will see that when there are 7 black
marbles and 3 white marbles, the probability of choosing one of each is 7/15. However, it would also be
true in reverse: If there were 7 white marbles and 3 black marbles, the probability would also be 7/15.
Therefore, this is not enough information.
Combining them does give us enough information. From statement 2 we know that there must be 7 of
one color and 3 of the other, and from statement 1 we know that there must be more black than white, so
we know there must be 7 black marbles and 3 white marbles.
Answer: C2. In a class with 12 children, q of the children are girls. Two children will be randomly chosen
simultaneously. What is the value of q?
1) The probability that two girls will be chosen together is 1/11.
2) The probability that one boy will be chosen and one girl will be chosen is 16/33.
A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though
NEITHER statement BY ITSELF is sufficient.
D) Either statement BY ITSELF is sufficient to answer the question.
E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that
further information would be needed to answer the question.SolutionTo answer this question, you can do the math, or you can rely on the experience you have gained thus far.
Let’s work out statement 1 by thinking it through:
Statement 1: We know there are a specific number of girls (q). Since each number of girls would yield a
different probability of choosing 2 girls, there must be only one specific number that would yield 1/11. So it
must be enough information.
Now, statement 2 requires a little more thought. Let’s work it out by doing the math:Statement 2: This one may seem to follow the same logic, as they are giving us a specific probability. However, this time we are asked to pick one boy and one girl. Look at the following chart to see why this isn’t enough information:
Boys Girls P(1 boy and 1 girl)*
1 11 1/12 × 11/11 = 1/6
2 10 2/12 × 10/11 = 10/33
3 9 3/12 × 9/11 = 9/22
4 8 4/12 × 8/11 = 16/33
5 7 5/12 ×7/11 = 35/66
6 6 6/12 × 6/11 = 6/11
7 5 7/12 × 5/11 = 35/66
8 4 8/12 × 4/11 = 16/33
9 3 9/12 × 3/11 = 9/22
10 2 10/12 × 2/11 = 10/33
11 1 11/12 × 1/11 = 1/6
*Note: we will multiply each probability by 2, because we can choose a boy and a girl, or a girl and a boy,
and both will yield the desired result.
As you can see, each probability is repeated for inverse combinations of boys and girls. There are two
ways to get 16/33, once with 4 boys and 8 girls, and also with 4 girls and 8 boys. This is not enough
information. We do not know what q is.
3. In a hotel with single rooms and double rooms, what is the probability that a room chosen at random will
be a double room painted red?
1) 1/6 of the rooms in the hotel are painted red.
2) 2/3 of the hotel’s rooms are double rooms.
A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though
NEITHER statement BY ITSELF is sufficient.
D) Either statement BY ITSELF is sufficient to answer the question.
E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that
further information would be needed to answer the question.
SolutionStatement 1 tells us the fraction of rooms painted red, but we do not learn anything about double rooms.
Statement 2 tells us the fraction of the rooms that are double rooms, but we do not know anything about
red rooms.
Putting the statements together still does not give us enough information because we do not know how
many of the red rooms are double rooms. Imagine if we said there were 12 rooms. We would then know
that 2 were red, and 8 were doubles. But we do not know if any of the doubles are red or not. There is
simply no information connecting the two categories, so we cannot solve the probability (E).
4. Two identical dice are rolled together. If the sum of the dice is 7, what is the probability that one of the
numbers showing is a 4?
A) 1/36
B) 1/18
C) 1/6
D) 11/36
E) 1/3
SolutionFor this problem, we want to calculate the probability of rolling a 4 knowing that the sum of the dice is 7.
We therefore do not have to take into account any other combinations of dice other than those that equal 7:
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
There are six pairs that equal seven, and 2 of them have a four in them. Therefore, the probability is 2/6 =
1/3.
Answer: E
5. At 3 pm, Jennifer went into labor. There is a 70% chance her baby will be born each hour that she is in
labor. What is the probability that her baby will be born at 6 pm on the same day?
A) .027
B) .063
C) .147
D) .27
E) .343
Solution
This question is an independent probability question in disguise. The probability that her baby will be born
each hour does not change. Each hour, there is a 70% chance the baby will be born, which means there is
a 30% chance the baby will not be born.
Therefore, we can see that from 3 pm – 4 pm, the baby is not born and the probability of that happening is
30%, from 4 pm – 5 pm, the probability is 30%, and from 5 pm – 6 pm, when the baby is born, the
probability is 70%. Since the baby must not be born in the first hour, and must not be born in the second
hour, and must be born in the third hour, the probability is (0.3)(0.3)(0.7) = .063.4.
Answer: B
6. A fair coin is to be flipped four times. What is the probability that the coin will land on the same side on
all four flips?
A) 1/32
B) 1/16
C) 1/8
D) 1/4
E) 1/2SolutionThe probability the coin will land on the same side is the probability that it will land on all heads or all tails.
All heads looks like: H H H H and all tails looks like: T T T T.
The probability of each is 1/2 × 1/2 × 1/2 × 1/2 = 1/16. Since either could happen, we will add the two
probabilities together: 1/16 + 1/16 = 2/16 = 1/8.Answer: C
The probability, combinations, permutations chapter is complete.