F. Permutations
I. Introduction
The first thing to know about the subject of permutations on the GMAT
is that it is simpler than most people think. These questions are
usually associated with the toughest questions on the test that signal
a high score. While that is generally true, the questions can be answered
by using a little logic and a lot of practice.
Before beginning, let’s look at a sample problem just to understand
what permutation questions are all about. We won’t try to solve
it now:
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?
As you can see, permutation questions are about taking a group of
objects and totaling how many ways we can arrange them in various
orders.
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II. The Basics: Three Steps to Permutation Clarity
There are three main steps to working out permutations questions:
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 1
How many outcomes are there when two identical dice are rolled?
Answer = 36
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_ x _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?
Answer = 160
Following the steps:
1. Figure out how many places there are to fill
Because there are three digits, there are three places to fill:
__ __ __
2. Figure out how many objects
potentially can go into each place
The question states that the first digit can be any number from 2-9,
inclusive. There are therefore 8 potential options. The second digit
can be only 0 or 1, therefore, there are 2 potential options. The
third digit can be any number from 0-9, inclusive, and there are 10
such numbers. The diagram looks like this:
_8_ _2_ _10_.
3. Multiply for the answer
_8_ x _2_ x _10_ = 160.
III. Permutations Without Repeating Values
In the previous two examples, values were allowed to repeat themselves.
For example, it is possible to roll double 5’s when rolling
two dice. This is different than cases where you don’t have
repeats, such as a single deck of cards where if four aces are pulled,
there are no aces left. In the next examples, we will apply the same
concepts, but ensure that no replacements or value repeats. Remember,
when doing these questions always think of the number of places to
fill in, and the potential number of options for each place.
Example 3
In how many ways can 5 people sit on a 5 person bench?
Answer = 120
In this case, we cannot have repeating values. If someone named Bob
is already sitting, he cannot appear again at a different place on
the bench! Let’s follow the steps, then, thinking about the
logic involved.
1. Figure out how many places there are to fill
There are 5 seats on the bench, so there are 5 places:
__ __ __ __ __
2. Figure out how many objects potentially can go into each
place
There are 5 people who could sit in the first seat. Once someone actually
does sit, though, there are only 4 people who could sit in the second
seat. For the third seat, there are only 3 people left, and then 2
and then 1. So the places will look like this:
_5_ _4_ _3_ _2_ _1_
3. Multiply for the answer
_5_ x _4_ x _3_ x _2_ x _1_ = 120
What is a Factorial?
By definition, 5 factorial (written as “5!”) means multiplying
5 by every number below it including 1. The above solution can be
written as “5 factorial” and the symbol for factorial
is the exclamation point. So the answer to example 3 is 5! (where
“!” is the symbol of factorial).
Example 4
There are six meal options in the cafeteria of a certain school. Assuming
that a different meal must be eaten each day, and each different type
of meal must be eaten once before any type of meal can be eaten a
second time, how many different ordering options
are there for a student in the first four days?
Solution
The answer is 360. In this case we cannot have repeating values not
because of the nature of the situation, but because we are told that
a student cannot eat two of the same types of meal before they go
through all the meals first.
Following the steps:
1. Figure out how many places there are to fill:
We want to know how many ordering options a student has on the first
four days, so we have four places to fill.
___ ___ ___ ___
2. Figure out how many objects
potentially can go into each place:
There are 6 meal options that a student
can order from on the first day. Since he/she cannot repeat a meal
before all are tried, there are only 5 meal options on the next day.
On day three, there are 4 options, and on day 4 three options.
_6_ _5_ _4_ _3_
3. Multiply for the answer
_6_x_5_x_4_x_3_ = 360
IV. GMAT Problem Variations
By now you should see that solving permutations problems is a simple,
three step process involving a system, a bit of logic, and confidence.
The GMAT, of course, thinks you should use that logic on more challenging
problems. But do not fear! The system and the logic do not change.
If you just remember to figure out the number of places and the potential
options for each one, you will do fine.
For our next example, let’s return to the first problem.
Example 5
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?
Answer = 36
1. Figure out how many places there are to fill
There are six cages for the animals, so there are six places:
__ __ __ __ __ __
2. Figure out how many objects potentially can go into each
place
This is where things change, but just a bit. There are 3 cats and
3 dogs, but the cats must be in the second, fourth, and sixth cages.
That means the dogs must be in the first, third, and fifth cages.
If we go cage by cage, we can figure our the potential number of options
for each cage (remember, there can be no repeating values):
_3_ _3_ _2_ _2_ _1_ _1_
3. Multiply for the answer
_3_ x _3_ x _2_ x _2_ x _1_ x _1_ = 36
Continue to next page, Combinations
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