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The number of arrangement of letters of the word BANANA in

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The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 28 Oct 2013, 05:26
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A
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Question Stats:

61% (01:44) correct 39% (01:51) wrong based on 240 sessions

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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
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Re: The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 28 Oct 2013, 07:03
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saurabhprashar wrote:
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


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.
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The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 14 Oct 2018, 05:31
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?
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Re: The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 14 Oct 2018, 09:20
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philipssonicare wrote:
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?


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

Cheers,
Brent
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Re: The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 14 Oct 2018, 09:30
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saurabhprashar wrote:
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


---------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
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Re: The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 28 Apr 2019, 05:47
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.
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Re: The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 28 Apr 2019, 05:49
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.
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Re: The number of arrangement of letters of the word BANANA in  [#permalink]

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New post 02 Jun 2019, 02:22
saurabhprashar wrote:
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


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
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Re: The number of arrangement of letters of the word BANANA in   [#permalink] 02 Jun 2019, 02:22
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