Last visit was: 15 Sep 2024, 19:39 It is currently 15 Sep 2024, 19:39
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 95518
Own Kudos [?]: 658968 [13]
Given Kudos: 87262
Intern
Joined: 05 Feb 2017
Posts: 26
Own Kudos [?]: 26 [5]
Given Kudos: 50
General Discussion
Senior Manager
Joined: 29 Jun 2017
Posts: 306
Own Kudos [?]: 827 [0]
Given Kudos: 76
GPA: 4
WE:Engineering (Transportation)
Intern
Joined: 05 Feb 2017
Posts: 26
Own Kudos [?]: 26 [0]
Given Kudos: 50
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
sahilvijay
Bunuel
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?

A. 112
B. 96
C. 72
D. 24
E. 6

IMO C

x1x3x5x
Four x places for 2 => 4C1
remaining three x places for 4 => 3C1
and 1 ,3,5 can be rotated in 3!

4C1.3C1.3! = 4x3x3! =12x6=72

Just a thought

Four x places for 2 => 4C1
remaining three x places for 4 => 3C1

We can write
4C2 * 2! because we have 4 places for 2,4 and they can be arranged in 2! ways.
Senior Manager
Joined: 29 Jun 2017
Posts: 306
Own Kudos [?]: 827 [1]
Given Kudos: 76
GPA: 4
WE:Engineering (Transportation)
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
1
Kudos
prerakgoel03
sahilvijay
Bunuel
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?

A. 112
B. 96
C. 72
D. 24
E. 6

IMO C

x1x3x5x
Four x places for 2 => 4C1
remaining three x places for 4 => 3C1
and 1 ,3,5 can be rotated in 3!

4C1.3C1.3! = 4x3x3! =12x6=72

Just a thought

Four x places for 2 => 4C1
remaining three x places for 4 => 3C1

We can write
4C2 * 2! because we have 4 places for 2,4 and they can be arranged in 2! ways.

just one more thought

attach 2 to 1 like 21 and attach 4 to 3 like 43 so it will be
21 43 5. 21 and 12 is 2 ways
21 43 and 5 are 3 units can be arranged in 3! = 6
and 2 can be attached to 1,3,5 =3
and 4 can be attached to 3,5 ( 1 less way than above) = 2
so total ways = 2 x 3! x 3 x 2 = 2 x 6 x 6 = 72

Another fantastic way to solve
What say--- kudo bante h ki nahi prerakgoel03
Senior Manager
Joined: 29 Jun 2017
Posts: 306
Own Kudos [?]: 827 [0]
Given Kudos: 76
GPA: 4
WE:Engineering (Transportation)
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
prerakgoel03 : i hope you have understood the latest method posted by me.
Senior Manager
Joined: 29 Jun 2017
Posts: 306
Own Kudos [?]: 827 [0]
Given Kudos: 76
GPA: 4
WE:Engineering (Transportation)
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
prerakgoel03
sahilvijay
Bunuel
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?

A. 112
B. 96
C. 72
D. 24
E. 6

IMO C

x1x3x5x
Four x places for 2 => 4C1
remaining three x places for 4 => 3C1
and 1 ,3,5 can be rotated in 3!

4C1.3C1.3! = 4x3x3! =12x6=72

Just a thought

Four x places for 2 => 4C1
remaining three x places for 4 => 3C1

We can write
4C2 * 2! because we have 4 places for 2,4 and they can be arranged in 2! ways.

-------------prerakgoel03 - one more method

-> total ways 1 2 3 4 5 = 5 ! = 120
number of ways in which 2,4 will be together = 2,4 is x
-1- 3- 5-
4C1 they can come
2! to arrange internal
3! for 135 arrrange
total to be subtracted = 4x2x3! = 8x6 = 48
120-48 = 72 Option C
Senior Manager
Joined: 29 Jun 2017
Posts: 306
Own Kudos [?]: 827 [0]
Given Kudos: 76
GPA: 4
WE:Engineering (Transportation)
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
prerakgoel03 one last way

-2-4-
fill 1,3,5 in these
1 can come in 3 ways followed by 3 in 2 ways and 5 in 1 way = 3x2x1 = 6
selecting 1 from 1,3,5 = 3c1
3 from remaining 3,5 = 2c1
and 5 = 5c1
2 and 4 can be arranged in 2!
total ways = 6x3c1 x 2c1 x 2 = 6x6x2 = 72 again C
Target Test Prep Representative
Joined: 04 Mar 2011
Affiliations: Target Test Prep
Posts: 3036
Own Kudos [?]: 6849 [2]
Given Kudos: 1646
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
1
Kudos
1
Bookmarks
Bunuel
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?

A. 112
B. 96
C. 72
D. 24
E. 6

We can use the following formula:

Total number of arrangements = # ways with 2 and 4 adjacent + # ways with 2 and 4 not adjacent

The total number of arrangements is 5! = 120

Next we can determine the number of arrangements when 2 and 4 are adjacent. We can make the numbers 2 and 4 one placeholder, such that there are 4 total positions or 4! = 24. However, we must include that we can arrange the 2 and 4 in 2! = 2 ways. So, the total number of ways is 24 x 2 = 48.

Therefore, the number of ways with 2 and 4 not adjacent is 120 - 48 = 72.

GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6803
Own Kudos [?]: 31327 [1]
Given Kudos: 799
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]
1
Kudos
Top Contributor
Bunuel
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?

A. 112
B. 96
C. 72
D. 24
E. 6

Take the task of seating the 5 digits and break it into stages.

Stage 1: Arrange the 1, 3 and 5
Since n unique objects can be arranged in n! ways, we can arrange these 3 digits in 3! (6) ways.
We can complete stage 1 in 6 ways.

Key step: For each arrangement of the 3 digits, add a space on each side of each digit. So, for example, if we add spaces to the arrangement 153, we get: _1_5_3_
Each of the 4 spaces represents a possible location for the two remaining digits (2 and 4). Notice that this configuration ensures that the 2 and 4 can't be adjacent.

Stage 2: Choose a space to place the 2.
There are 4 available spaces, so we can complete stage 2 in 4 ways.

Stage 3: Choose a space to place the 4.
There are 3 available spaces remaining, so we can complete stage 3 in 3 ways.

At this point, we'll throw away the remaining spaces, leaving an arrangement with all 5 digits.

By the Fundamental Counting Principle (FCP), the number of ways to complete all 4 stages (and thus place all 8 letters) = (6)(4)(3) = 72 ways