You could approach it by viewing it as three decisions and then multiplying the
number of options for each decision together. There are 6 candidates for the first
song, which leaves 5 possibilities for the second song and 4 possibilities for the
third song. That creates 6
×
5
×
4 = 120 possible versions of the CD.
Assignment 6
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191
Alternatively, you could use the permutation formula to get the
same result. There are 6 candidates, so
n
= 6. Elmo is choosing 3 songs, so
r
= 3.
When
you
plug
these
numbers
into
the
formula,
you
get
P
n
n
r
=
−
=
=
×
×
×
×
×
×
×
=
!
(
)!
!
!
6
3
6
5
4
3
2
1
3
2
1
120
.
A third possibility is to break this problem into two subproblems: a
combination and a permutation. It’s not the most efficient way to answer the
questions but it will provide some insight into the way combinations and
permutations fit together. This understanding can be critical for answering more
difficult questions that do not fit nicely into the patterns you’ve seen up to now.
You can analyze Elmo’s decision in two stages:
1.
He must decide which 3 songs he wants to include. This is a
combination because order is not involved.
2.
Once he has chosen the 3 songs, he must decide in what order to
record them. This is a permutation because order is involved.
In the first phase, you find the number of 3song groups (combinations) that
are possible. Then, you multiply by the number of ways to arrange a given
3song group. That will give you the total number of permutations.
# of combinations
of 3 songs
# of ways to order a
group of 3 songs
Total # of
pemutations
×
=
The first step is to determine how many 3song combinations are possible.
You could use the group formula for this:
C
n
r
n
r
=
−
=
=
×
×
×
×
×
×
×
×
×
×
=
!
!(
)!
!
! !
6
3 3
6
5
4
3
2
1
3
2
1
3
2
1
20
. The 20 possible groups include
ABC, ABD, ABE, and so forth.
Next, suppose that Elmo has selected his 3 most favorite songs. He needs to
determine the order in which to record these songs on the CD. This a permutation
in which you choose 3 out of 3 items and arrange them. There are 3
×
2
×
1 = 6
choices. You could also use the permutation formula, with
n
= 3 and
r
= 3. You
will get the same result. (Remember that 0! = 1.)
So, there are 20 possible groups (combinations), each of which could be
arranged in 6 different ways, for a total of 20
×
6 = 120 permutations.
The key insight about combinations and permutations is that a permutation
is just a combination that has been arranged in some order. Conversely, a
combination is just a permutation with the ordering factored out. That’s why
there is an extra
r
! in the denominator of the combination formula.
P
n
n
r
=
−
!
(
)!
C
n
r
n
r
P
r
=
−
=
!
!(
)!
!
GMAT MANUAL
192

© Princeton Review Management, L. L. C.
C
IRCULAR
P
ERMUTATIONS
Questions involving circular permutations are very rare. In fact, it’s fine to skip
this section. It’s only worth spending time on this topic if you’ve mastered
everything else.
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