G. Combinations
Combinations problems are very similar to permutation problems. The
key distinction is that order does not matter for combinations, and
in permutations it does.
For example, if a committee is being
put together for a company and there is a president, vice president,
and treasurer, order matters. If that same committee is being
put together but nobody has a rank then order does not matter
because there is no hierarchy.
I. Example of Combinations
Let's say there is a group of 4 children
(Albert, Becky, Chuck, and Devin) who must be put on a three person
line to go to an Art Class. How many different ways can this be arranged?
Firstly, before we solve, ask yourself
the question, "Does order matter in this problem?" The answer
is yes, a line is something where the placement of a person would
matter. So, this is a permutation problem.
1. Figure out how many places
there are to fill
This is a 3 person line, so there are three spaces to fill:
__ __ __
2. Figure out how many objects
potentially can go into each place
In the first space there are 4 children to choose from, then only
3 to choose from for the second space, and 2 for the third.
_4_ _3_ _2_
3. Multiply for the answer
_4_x_3_x_2_ = 24
Here is a chart of all the possible
permutations:
Spot1 Spot2 Spot3 Spot1
Spot2 Spot3 Spot1
Spot2 Spot3 Spot1
Spot2 Spot3
Albert Becky Chuck Becky Albert Chuck
Chuck Albert Becky Devin Albert Becky
Albert Chuck Becky Becky Chuck Albert
Chuck Becky Albert Devin Becky Albert
Albert Chuck Devin Becky Albert Devin
Chuck Albert Devin Devin Chuck Albert
Albert Devin Chuck Becky Albert Devin
Chuck Devin Albert Devin Albert Chuck
Albert Becky Devin Becky Chuck Devin
Chuck Becky Devin Devin Becky Chuck
Albert Devin Becky Becky Devin Chuck
Chuck Devin Becky Devin Chuck Becky
Here you can clearly see how all the
students can be set up in 24 different ways.
Now let's do the combinations version
of this question.
Let's say that the same group of 4 children (Albert, Becky, Chuck,
and Devin)must be picked to be put in an Art Class. How many different
ways can this be arranged?
Once again, we ask the question, "Does order matter?" In
this case it does not matter because when the children are
put into a class, there is no order placed on them. It would not matter
if it was Devin, Becky, and Chuck in a class or Becky, Devin and Chuck
because when you look into the classroom you would see the same exact
thing.
So, let's look at our chart from the last problem and identify which
cases we would now not count because the order does not matter:
Spot1 Spot2 Spot3 Spot1
Spot2 Spot3 Spot1
Spot2 Spot3 Spot1
Spot2 Spot3
Albert Becky Chuck Becky
Albert Chuck Chuck Albert Becky
Devin Albert Becky
Albert Chuck Becky
Becky Chuck Albert Chuck
Becky Albert Devin Becky Albert
Albert Chuck Devin Becky
Devin Albert Chuck Albert Devin
Devin Chuck Albert
Albert Devin Chuck
Becky Albert Devin Chuck
Devin Albert Devin Albert Chuck
Albert Becky Devin Becky Chuck Devin
Chuck Becky Devin Devin Becky Chuck
Albert Devin Becky
Becky Devin Chuck Chuck
Devin Becky Devin Chuck Becky
If we count the final results, we get only 4 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.
1. Do the problem as if it was a permutations problem.
2. Divide the answer by (the number of spaces)! (factorial)
Let's apply this to our line/class problem.
1. Do the problem as if it was a permutations problem.
We have the answer from the previous problem for how to do it as
a permutations problem, and as you remember there 24 possibilities.
2. Divide the answer by (the number of spaces)! (factorial)
Now we will divide 24 by the number of spaces factorial, or 3!.
3! = 3 x 2 x 1 = 6
Therefore 24 / 6 = 4, which is the total amount of combinations of
these 4 children into 3 person classes there can be.
800Score.com 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.
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.
II. Is it a Permutation or Combinations question?
As we stated above, a major part of the challenge on GMAT questions
is to determine if a question is permutations or combinations. To
do this, look for some important clues:
1. Does order matter? This is the most important question, where
you should ask yourself after having read a question “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 word “permutations”
will not be in a GMAT problem.
3. If the word “combinations” is
in the problem, it is a combinations problem.
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.
//Insert the table from Word here.
III. GMAT Problem Variations
Now that you understand combinations problems and the method to solve
them, there are two important combinations question types 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 8//??????
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
a problem which would just simply ask how many different combinations
of the toppings there could be on a 10 topping pizza. This is because
there are many ways Jonathan 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
_8_ x _7_ x _6_ x _5_
Vegetables
_7_x_6_ x _5_
Cheeses
_5_ x _4_ x _3_
2. Divide the answer by (the number of spaces)!
Meats
_8_ x _7_ x _6_ x _5_ / 4 x 3 x 2 x 1
Vegetables
_7_x_6_ x _5_ / 3 x 2 x 1
Cheeses
_5_ x _4_ x _3_ / 3 x 2 x 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
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 x 5 / 2 x 1 = 15
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