GMAT
COMBINATIONS
Combinations problems are very similar to permutation problems. The key distinction is that
placement/order isn’t relevant for combinations, but placement/order is relevant for permutations.
Here’s an example:
Let’s say we play a game of dice where we roll two dice (one red and one blue) and record the results.
If the dice roll up the same number, the results aren’t counted and we roll again. So there are no
doubles.
How many permutations are possible?
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
Since you cannot get the same result (doubles) on the second die there are only 5 possibilities
for the second roll.
_6_ _5_.
3. Multiply for the answer
_6_ × _5_ = 30
red blue
1 1
1 2
1 3
1 4
1 5
1 6
red blue
2
1
2
2
2
3
2
4
2
5
2
6
red
3
3
3
3
3
3
blue
1
2
3
4
5
6
red blue
4
1
4
2
4
3
4
4
4
5
4
6
red blue
5
1
5
2
5
3
5
4
5
5
5
6
red
6
6
6
6
6
6
blue
1
2
3
4
5
6
You eliminate all doubles (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.
Now let’s do the “combinations” versions of this 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 is a combinations question and not a permutations question.
Strategic tip: half of the challenge with most combination/permutation questions is correctly
identifying if it is a combination or permutation question. Once you know this, most medium and lower
skill level GMAT questions are pretty simple.
red red
1 1
1 2
1 3
1 4
1 5
1 6
red red
2
2
2
2
2
2
red
1
2
3
4
5
6
red
3
3
3
3
3
3
1
2
3
4
5
6
red red
4
1
4
2
4
3
4
4
4
5
4
6
red red
5
1
5
2
5
3
5
4
5
5
5
6
red
red
6
6
6
6
6
6
1
2
3
4
5
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) when the two red dice will look the same. So because
this is a combinations problem, we have to reduce the total number of permutations. All the
possibilities that are overlaps we will mark as green as we move from left to right. So in the second
column (2,1) is marked green because it is the same as (1,2). In the third column (3,1)(3,2) are both
marked green because they are copies of (1,3) and (2,3).
red red
1 1
1 2
1 3
1 4
1 5
1 6
red red
2
2
2
2
2
2
red
1
2
3
4
5
6
Red
3
1
3
2
3
3
3
4
3
5
3
6
red red
4
1
4
2
4
3
4
4
4
5
4
6
red red
5
1
5
2
5
3
5
4
5
5
5
6
red
red
6
6
6
6
6
6
1
2
3
4
5
6
If we count the final results, we get 15 total combinations. Obviously you can’t rely on using charts
like this on test day, so there must be a simpler way to solve these combinations problems.
Once you figure out that a problem is a combinations problem, you need to only follow two steps:
1. Do the problem as if it was a permutations problem.
2. Divide the answer by (the number of spaces)! (factorial)
Or, you can combine those two steps into this simple combinations formula:
combinations =
permutations
(# of spaces)!
Let’s apply this to our dice problem. We have two dice, so there are two spaces. Following the
combinations formula:
combinations =
65 65

 3  5  15
2!
2 1
Notice that there are only 15 combinations while there are 30 permutations.
Common-Sense Tip: 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.
So if there are 60 permutations, and 3 spaces, you end up just getting 10 combinations. Combinations
are ALWAYS smaller so long as the # of spaces is greater than 1. It is logical that there are fewer
combinations than permutations because combinations don’t distinguish between order and placement,
so there are fewer possibilities.
As we stated above, a major part of the challenge is to determine if a question is permutations or
combinations. To do this, look for some important clues:
1. If the word “arrangements” is in the problem, it is a permutations problem. The word
“permutations” will not be in a GMAT problem.
2. If the word “combinations” is in the problem, it is a combinations problem.
3. If the number of places and the number of objects is the same (why?)
4. Use your logic and common sense.
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 permutations and combinations.







Permutations
Combination of a safe or lock
Different numbers (i.e., ID Codes, or
number of 4-digit numbers, etc)
Committees with position titles
Order of winning 1st, 2nd, 3rd in a race
(also Gold, Silver, Bronze).
Arranging people in a row.
Different words to be made from the
same letters.
Different outfits to wear from a full
wardrobe.







Combinations
Generic committees
Different colors used in a painting or
project.
Different materials used in a project.
Different clothes to take on vacation
from a larger selection.
Different prizes to take home from a
larger selection.
Different ingredients to be mixed
together from a larger selection.
Handshakes.
Example 8
There are 15 available toppings in a pizza restaurant. If Maria will order a 4-topping pizza, how
many different pizzas could she order?
Answer = 1365
As we’ve already seen, the order of toppings on a pizza does not matter, so this is a combinations
question. We want to eliminate all the redundant pizzas and only count the different ones.
Two Steps to Combinations:
1. Do the problem as if it was a permutations problem.
a. Figure out how many places there are to fill
There are 4 spaces because there are four toppings: __ __ __ __
b. Figure out how many objects potentially can go in each place
_15_ _14_ _13_ _12_
c. Multiply
_15_ x _14_ x _13_ x _12_
Time-Saving Combinations Tip: This time, don’t do the multiplication – we will be dividing and
that’ll save time!
2. Divide the answer by (the number of spaces)!
There are 4 spaces, so that’s 4! That looks like this:
15  14  13  12 15  14
  7  13  12


 15  7  13  1365
4  3 2 1
4  3  2  1
This may look like a lot in the beginning, but you will soon see that it is a natural and quite simple
operation, which works across the board for all combinations questions.
Example 9
An abstract painter has 15 colors on her pallet to work with. If she decides she will paint with exactly
5 colors, how many different combinations of colors can she choose from?
Answer = 3003
1. Do the problem as if it was a permutations problem (remember NOT to multiply!).
a. Figure out how many spaces there are to fill
There are 5 colors so there are 5 spaces: __ __ __ __ __
b. Figure out how many objects potentially go in each place
_15_ _14_ _13_ _12_ _11_
c. Multiply:
_15_ x _14_ x _13_ x _12_ x _11_
2. Divide the answer by (the number of spaces)!
There are five spaces, so divide by 5!
15  14  13  12  11
 3003
5 4  3 2 1
II. GMAT Problem Variations
Now that you understand combinations problems and the method to solve them, there are two
important combinations question variations you should be aware of when preparing for the exam.
Variation 1: Combinations from Multiple Groups
In this situation, combinations are being drawn from several groups to form a complete set. Figure out
the combinations from each group and then multiply them together.
Example 10
At Sam’s Pizza Parlor, there are 8 meats, 7 vegetables, and 5 cheeses to choose from. Jonathan
would like to make a pizza with 4 meats, 3 vegetables, and 3 cheeses. How many different pizzas
could he order?
Answer = 24,500
This problem is different from example 8. In example 8, Maria wanted to order a pizza mixing all the
toppings together. In this case, Jonathan wants his pizza a specific way. But there are many ways he
can have meat on his pizza, many ways he can have vegetables, and many ways he can have cheese.
To answer this correctly, you need to solve each combinations problem individually and then multiply
the answers together for the correct answer.
1. Do the problem as if it was a permutations problem.
Meats
Vegetables
Cheeses
_8_ x _7_ x _6_ x _5_ _7_ x _6_ x _5_
_5_ x _4_ x _3_
2. Divide the answer by (the number of spaces)!
Meats
Vegetables
Cheeses
8765
4  3 2 1
765
3 2 1
54 3
3 2 1
Remember to cancel before you multiply to make it easier!
Meats
Vegetables
Cheeses
70
35
10
3. Multiply the answers together for the final answer.
70 x 35 x 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?
Answer = 5600
1. Do the problem as if it was a permutations problem.
Green
Red
Purple
_6_ x _5_ x _4_
_8_ x _7_ x _6_
_5_
2. Divide the answer by (the number of spaces)!
Green
Red
Purple
654
3 2 1
876
3 2 1
5
1
Remember to cancel before you multiply to make it easier!
Green
Red
Purple
20
56
5
3. Multiply the answers together for the final answer.
20 x 56 x 5 = 5600
Variation 2: Pairings
Sometimes, the simplest questions are the most difficult. Combinations questions with only 2 spaces
can be very tricky, but with a little practice you’ll recognize them and tackle them with ease.
Remember, 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?
Answer = 15
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!
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_ x _5_
2. Divide the answer by (the number of spaces)!
There are two spaces, so divide by 2!
65
 15
2 1
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?
Answer = 525
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 can 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_ x _30_
2. Divide the answer by (the number of spaces)!
There are two spaces, so divide by 2! And don’t forget to cancel!
35  30
 35  15  525
2 1