In how many ways can we rearrange the letters of the word MANANA such

Question

In how many ways can we rearrange the letters of the word MANANA such that no two A’s are adjacent to each other?

A. 12
B. 18
C. 24
D. 30
E. 60

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shameekv1989 · Answer

In how many ways can we rearrange the letters of the word MANANA such that no two A’s are adjacent to each other?

A. 12
B. 18
C. 24
D. 30
E. 60

Arrange letters in this way :- 
_M_N_N_

Select any 3 places for 3 A's among the 4 _ in the above.

Total Combinations :- 
 \frac{3!}{2!} * 4_C_3  = 3*4 = 12 ways

Answer - A

ananya32 · Answer

Place all A's in the given fixed place.
Only move around M and N's. 
 
So, for the word 
 M   A   N   A   N   A 
M and N's are at 3 places which can be arranged in 3! ways.

The last step is to multiply our result by 2 since the A's could be interchanged and fixed at the place of M and N's.

Ans: 3! x 2 = 12. Option (A)

Kritisood · Answer

Ans= total ways - whenever AAs are adjacent

total ways = 6!/3!2!= 60
2 AAs adjacent = 5!/2! = 30
3 AAAs adjacent. = 4!/2! = 12 
total when As are adjacent = 42 - i understand here there must be overlapping cases, how to understand those and subtract the same?

GMATinsight · Answer

Kritisood

Please note that 2AA's together is not 30

ATLEAST 2 AA adjacent will always include 3 AAA adjacent and the ordering of AAA might also come in picture so that's the problem

So, It's best to make two separate cases

1) Exactly 2A adjacent - MANANA has 3 As and 6 letters in total

AA - - - - now third A can come only on last three places and the remaining three letters can be arranged in 3!/2! ways (2N are identical) = 3*3!/2! = 9

- AA - - - now third A can come only on last two places and the remaining three letters can be arranged in 3!/2! ways (2N are identical) = 2*3!/2! = 6

- - AA - - now third A can come only on two places (first and last) and the remaining three letters can be arranged in 3!/2! ways (2N are identical) = 2*3!/2! = 6

- - - AA - now third A can come only on First two places and the remaining three letters can be arranged in 3!/2! ways (2N are identical) = 2*3!/2! = 6

- - - - AA now third A can come only on First three places and the remaining three letters can be arranged in 3!/2! ways (2N are identical) = 3*3!/2! = 9

2) Exactly 3A adjacent

AAA --- Total arrangements will be 4 (places for AAA) * 3!/2! = 12

Final Result = 6!/(3!2!) = 60 - (9+6+6+6+9) - 12 = 12

Answer: Option A

But I won't call it the best method. It's just a method to understand the dynamics of arrangements

Codebug4it · Answer

Answer:  Option A

Word is MANANA and no two A's should be together, so we will keep A fixed with a place between them:

_A_A_A_

=> M and N can be placed around A's.

=> 4 places are empty so we can arrange M and N in  4! ways

=> As N is repeated, so the total number of ways =   4! / 2!   =  12 (A)

ScottTargetTestPrep · Answer

We see that the 3 A’s can be separated (so that no 2 A’s are adjacent to each other) as follows:

A _ A _ A _

A _ _ A _ A

A _ A _ _ A

_ A _ A _ A

And the 3 empty spots are filled by 1 M and 2 N’s. For each of the 4 cases above, the 1 M and 2 N’s can fill the 3 empty spots in 3!/2! = 3 ways. Therefore, there are a total of 4 x 3 = 12 ways the letters of the word MANANA can be arranged such that no two A’s are adjacent to each other.

Answer: A

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In how many ways can we rearrange the letters of the word MANANA such

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