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

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Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129172 [0], given: 12194

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

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New post 11 Sep 2017, 05:54
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

62% (00:49) correct 38% (00:54) wrong based on 85 sessions

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Kudos [?]: 129172 [0], given: 12194

Senior Manager
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Joined: 29 Jun 2017
Posts: 345

Kudos [?]: 66 [0], given: 64

WE: Engineering (Transportation)
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How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 11 Sep 2017, 06:43
Bunuel wrote:
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

Answer choice C
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Kudos [?]: 66 [0], given: 64

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Kudos [?]: 9 [0], given: 49

Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 11 Sep 2017, 11:18
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IMO:-C

Total - When both are adjacent to each other

Total= 5!
Adjacent = 4!*2! (Make them a group so remaining 3 and a group can be arranged in 4! ways and the inside group in 2! ways)

Answer:- 5!-4!*2! = 120-48 = 72

Kudos [?]: 9 [0], given: 49

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Joined: 05 Feb 2017
Posts: 26

Kudos [?]: 9 [0], given: 49

Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 11 Sep 2017, 11:22
sahilvijay wrote:
Bunuel wrote:
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

Answer choice C


Just a thought
Instead of

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.

Kudos [?]: 9 [0], given: 49

1 KUDOS received
Senior Manager
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Kudos [?]: 66 [1], given: 64

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Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 11 Sep 2017, 11:35
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KUDOS
prerakgoel03 wrote:
sahilvijay wrote:
Bunuel wrote:
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

Answer choice C


Just a thought
Instead of

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
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Kudos [?]: 66 [1], given: 64

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Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 11 Sep 2017, 11:42
prerakgoel03 : i hope you have understood the latest method posted by me.
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Give Kudos for correct answer and/or if you like the solution.

Kudos [?]: 66 [0], given: 64

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Kudos [?]: 66 [0], given: 64

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

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New post 11 Sep 2017, 11:50
prerakgoel03 wrote:
sahilvijay wrote:
Bunuel wrote:
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

Answer choice C


Just a thought
Instead of

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 8-)
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Give Kudos for correct answer and/or if you like the solution.

Kudos [?]: 66 [0], given: 64

Senior Manager
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Joined: 29 Jun 2017
Posts: 345

Kudos [?]: 66 [0], given: 64

WE: Engineering (Transportation)
GMAT ToolKit User Reviews Badge
Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 11 Sep 2017, 11:58
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
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Kudos [?]: 66 [0], given: 64

Expert Post
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Kudos [?]: 837 [0], given: 5

Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such [#permalink]

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New post 14 Sep 2017, 10:35
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Bunuel wrote:
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.

Answer: C
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Kudos [?]: 837 [0], given: 5

Re: How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such   [#permalink] 14 Sep 2017, 10:35
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