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How many different arrangements of the digits 1, 2, 3, 4, and 5 are po

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Math Expert
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V
Joined: 02 Sep 2009
Posts: 47978
How many different arrangements of the digits 1, 2, 3, 4, and 5 are po  [#permalink]

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New post 25 Dec 2017, 01:51
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

42% (00:44) correct 58% (02:10) wrong based on 19 sessions

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Joined: 07 Dec 2017
Posts: 553
Re: How many different arrangements of the digits 1, 2, 3, 4, and 5 are po  [#permalink]

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New post 25 Dec 2017, 03:26
Bunuel wrote:
How many different arrangements of the digits 1, 2, 3, 4, and 5 are possible where no two consecutive digits are adjacent?

A. 42
B. 21
C. 14
D. 10
E. 6


As it is hard to calculate this directly, we'll start by writing down numbers and see how it goes.
This is an Alternative approach.

If we start with a 1 then the next number must be 3,4 or 5.
Starting with 1-3 means that 5 must come next (as 2 and 4 are adjacent to 3) then 2 then 4.
Similarly, 1-4-2-5-3 must follow from 1-4.
However, if we try to continue 1-5-3 we become stuck as both 4 and 2 are impossible.
So, there are 2 numbers starting with 1.
Let's start with 2:
2 can be followed by 4 or 5. Building these gives
2-4-1-3-5 and 2-4-1-5-3
2-5-1- impossible! and 2-5-3-1-4
We have a 3 numbers starting with 2.
Moving on to 3, which can be followed by 1 or 5:
3-1-4-2-5 and 3-1-5-2-4
3-5-2-4-1 and 3-5-1-4-2
4 more options.
Finally, we can list another 3 numbers for 4 and 2 for 5.
In total, we have 2+3+4+3+2=14 options.
(C) is our answer.

** Note: as 1,2,3,4,5 is a symmetrical sequence (5 and 1 are 'end numbers' and 2 and 4 are 'inbetween numbers') then we know that 1 and 5 have the same number of sequences without listing them. Similarly, 2 and 4 have the same number of sequences.
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Re: How many different arrangements of the digits 1, 2, 3, 4, and 5 are po  [#permalink]

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New post 25 Dec 2017, 06:43
Bunuel wrote:
How many different arrangements of the digits 1, 2, 3, 4, and 5 are possible where no two consecutive digits are adjacent?

A. 42
B. 21
C. 14
D. 10
E. 6


I had to list down all the cases, keeping the following conditions in mind.

    • 1 can be grouped with 3,4 and 5
    • 2 can be grouped with 4 and 5
    • 3 can be grouped with 1 and 5
    • 4 can be grouped with 1 and 2
    • 5 can be grouped with 1,2 and 3

I started of with 1 as the starting digit and found all the possible cases, then moved on to numbers starting with 2, 3, 4 and 5.

After writing down the cases, one can clearly see that the numbers formed with starting digits as 1,2 or 3, the number of cases are roughly 2 or 3 or 4 and thus, the answer would be somewhere between 10 and 16.

One may mark the answer as 14 and move on to the next question, if solving the question takes more than 2.5-3 minutes. :)

Total cases :

    • 1 3 5 2 4
    • 1 4 2 5 3
    • 2 4 1 3 5
    • 2 4 1 5 3
    • 2 5 3 1 4
    • 3 1 4 2 5
    • 3 1 5 2 4
    • 3 5 1 4 2
    • 3 5 2 4 1
    • 4 2 5 3 1
    • 4 2 5 1 3
    • 4 1 3 5 2
    • 5 2 4 1 3
    • 5 3 1 4 2

Thus, the correct answer is Option C.


Thanks,
Saquib
Quant Expert
e-GMAT
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Re: How many different arrangements of the digits 1, 2, 3, 4, and 5 are po &nbs [#permalink] 25 Dec 2017, 06:43
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