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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 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
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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
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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
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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
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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]
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
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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]
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]
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
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Re: The number of arrangement of letters of the word BANANA in
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YassiHASHMI wrote:
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.
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.
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Re: The number of arrangement of letters of the word BANANA in
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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.
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Re: The number of arrangement of letters of the word BANANA in [#permalink]
arrangement = 6!/2!*3!= 60 ways
and NNAAAB
NN=X
XAAAB = 5!/3! = 20 ways
totaal N is together ; 60-20 ; 40 ways
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Re: The number of arrangement of letters of the word BANANA in [#permalink]
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