The number of arrangement of letters of the word BANANA in

Question

The number of arrangement of letters of the word BANANA in which the two N's do not appear adjacently is:

A. 40
B. 50
C. 60
D. 80
E. 100

Bunuel · Accepted Answer

Total # of arrangements of BANANA is 6!/(3!2!) = 60 (arrangement of 6 letters {B}, {A}, {N}, {A}, {N}, {A}, where 3 A's and 2 N's are identical).

The # of arrangements in which the two N's ARE together is 5!/3!=20 (arrangement of 5 units letters {B}, {A}, {A}, {A}, {NN}, where 3 A's are identical)..

The # of arrangements in which the two N's do not appear adjacently is {total} - {restriction} = 60 - 20 = 40.

Answer: A.

BrentGMATPrepNow · Answer

---------ASIDE-------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are  n  objects where  A  of them are alike, another  B  of them are alike, another  C  of them are alike, and so on, then the total number of possible arrangements =  n !/ A !)( B !)( C !)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows: 
There are  11  letters in total 
There are  4  identical I's
There are  4  identical S's
There are  2  identical P's
So, the total number of possible arrangements =  11 !/ 4 !)( 4 !)( 2 !)]
-------ONTO THE QUESTION!!---------------------------------

In BANANA, there are: 
There are  6  letters in total 
There are  3  identical A's
There are  2  identical N's
So, the total number of possible arrangements =  6 !/ 3 !)( 2 !)]
=  60

IMPORTANT: Among these  60  outcomes, there are some outcomes that break the rule about N's not appearing next to each other.
So, let's determine the number of outcomes in which the 2 N's ARE together, and we'll subtract this from our  60  outcomes.

First "glue" the 2 N's together, to get  one  "super letter" NN
So, we now must arrange 5 letters: B, A, A, A, and NN
There are  5  letters in total 
There are  3  identical A's
So, the total number of possible arrangements =  5 !/( 3 !)
=  20

Number of arrangements that adhere to the rule =  60  -  20  = 40

Answer: A

Cheers,
Brent

philipssonicare · Answer

Bunuel 
These types of multinomial questions are great.
Is there a name for when two objects can't be adjacent to each other in combinations/permutations? It's a rule that I always forget how to apply.
A long shot, but do you know the source for OPs question?

BrentGMATPrepNow · Answer

I call these questions MISSISSIPPI questions. 
To see why, see my post below

Cheers,
Brent

YassiHASHMI · Answer

Hi Bunuel, 
Could you please confirm that while calculating restriction why we do not take into account reciprocal of {N1N2} i.e.{N2N1}? 
Is that because the reciprocal/repetition has already been excluded from the total combinations (by dividing 2!)? 
Thanks.

YassiHASHMI · Answer

Hi  Bunuel  
Could you please confirm that while calculating restriction why we do not take into account reciprocal of {N1N2} i.e.{N2N1}? 
Is that because the reciprocal/repetition has already been excluded from the total combinations (by dividing 2!)? 
Thanks.

Archit3110 · Answer

BANANA =
arrangement = 6!/2!*3!= 60 ways
and NNAAAB
NN=X
XAAAB = 5!/3! = 20 ways
totaal N is together ; 60-20 ; 40 ways
IMO A

MathRevolution · Answer

Total number of ways to arrange the letters of the given word BANANA :  \frac{(6!) }{ (3! * 2!)}  = 60

Number of ways if both NN are adjacent: Consider 'NN' as 1 letter: BAAA (NN)

=> We have 5 letters to arrange, therefore: 5!.

=> Both 'NN' can be arranged it 2! ways. As we have repetition (3A's and 2N's) therefore:

=>  \frac{(5!) *(2!) }{ (3! * 2!)}  = 20

The number of ways 'NN' won't be adjacent to each other: 60 - 20 = 40.

Answer A

aviddd · Answer

We don't consider N2N1 as a different arrangement since the two letters "N" are identical. There is no difference in the arrangement if you swap the positions of N.

Hope it is clear.

CEdward · Answer

6! / 3! x 2! = 60 <--- Total # of permutations with repeats
5! / 3! = 20 <--- Total # of permutations where Ns are TOGETHER

60 - 20 = 40 <--- Ps where they are not together

Answer is A.

gmatpro15 · Answer

arrangement = 6!/2!*3!= 60 ways
and NNAAAB
NN=X
XAAAB = 5!/3! = 20 ways
totaal N is together ; 60-20 ; 40 ways

Posted from my mobile device

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