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

Question

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

prerakgoel03 · Accepted Answer

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

sahilvijay · Answer

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

prerakgoel03 · Answer

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.

sahilvijay · Answer

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

sahilvijay · Answer

prerakgoel03  : i hope you have understood the latest method posted by me.

sahilvijay · Answer

------------- 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

sahilvijay · Answer

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

JeffTargetTestPrep · Answer

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

BrentGMATPrepNow · Answer

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

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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