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How many different arrangements of letters are possible if

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How many different arrangements of letters are possible if [#permalink]

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New post 27 Nov 2010, 07:49
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How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6
[Reveal] Spoiler: OA

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How many different arrangements of letters are possible if [#permalink]

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New post 27 Nov 2010, 07:57
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rxs0005 wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


Since A and E must be among 3 letters then the third letter must be out of B, C and D. 3C1 = 3 ways to choose which one it'll be. Now, 3 different letters (A, E and the third one) can be arranged in 3!=6 ways, so the final answer is 3*6=18.

Answer: D.
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Re: counting [#permalink]

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New post 03 Dec 2010, 04:01
Bunuel wrote:
rxs0005 wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


As A and E must be among 3 letters than the third letter must be out of B, C and D. 3C1=3 ways to choose which one it'll be. Now, 3 different letters can be arranged in 3!=6 ways, so final answer is 3*6=18.

Answer: D.


I could understand the first part that 3C1 , why cant we have 5C2*3C1

I sometimes fail to understand the basic diff when to apply permutation and when combination ?

if you can give a brief difference... thanks in advance
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Re: counting [#permalink]

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New post 03 Dec 2010, 05:10
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hirendhanak wrote:
Bunuel wrote:
rxs0005 wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


As A and E must be among 3 letters than the third letter must be out of B, C and D. 3C1=3 ways to choose which one it'll be. Now, 3 different letters can be arranged in 3!=6 ways, so final answer is 3*6=18.

Answer: D.


I could understand the first part that 3C1 , why cant we have 5C2*3C1

I sometimes fail to understand the basic diff when to apply permutation and when combination ?

if you can give a brief difference... thanks in advance


We are asked about the # of arrangements of 3 letters: {ABE} is a different arrangement from {EBA}, so for every group of 3 letters (for every selection of 3 letters) there will be 3 different arrangements possible and as there are total of 3 groups (3 selections) possible then there will be total of 3*6=18 arrangements.

Generally:
The words "Permutation" and "Arrangement" are synonymous and can be used interchangeably.
The words "Combination" and "Selection" are synonymous and can be used interchangeably.

Hope it's clear.
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Re: How many different arrangements of letters are possible if [#permalink]

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Re: How many different arrangements of letters are possible if [#permalink]

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Re: How many different arrangements of letters are possible if [#permalink]

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New post 13 Jul 2015, 11:51
Bunuel wrote:
rxs0005 wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


As A and E must be among 3 letters than the third letter must be out of B, C and D. 3C1=3 ways to choose which one it'll be. Now, 3 different letters can be arranged in 3!=6 ways, so final answer is 3*6=18.

Answer: D.


I got a bit tripped up in the wording here. I made the assumption that you could choose the same letter twice and got myself all sorts of confused. But after reading the OA it makes a lot of sense and hope I don't make these sorts of stupid mistakes in the future... sigh
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Re: How many different arrangements of letters are possible if [#permalink]

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New post 20 Jul 2015, 22:22
How do we know that we can't use the same letter twice?
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Re: How many different arrangements of letters are possible if [#permalink]

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New post 21 Jul 2015, 23:04
Bunuel wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


As A and E must be among 3 letters than the third letter must be out of B, C and D. 3C1=3 ways to choose which one it'll be. Now, 3 different letters can be arranged in 3!=6 ways, so final answer is 3*6=18.

Answer: D.[/quote]

Could we also solve this with: Total Combinations - Forbidden Combinations?

Total = 5*4*3 = 60
Forbidden (A is not part): 4*3*2 = 24
Forbidden (B is not part): 4*3*2 = 24
Forbidden (A and B are not part): 3*2*1 = 6

Total Forbidden Combinations: 54, Answer 6

I know its wrong but where is my mistake?
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Re: How many different arrangements of letters are possible if [#permalink]

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New post 19 Aug 2015, 13:06
Aves wrote:
How do we know that we can't use the same letter twice?


Yes, can someone answer this?
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Re: How many different arrangements of letters are possible if [#permalink]

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New post 19 Aug 2015, 13:19
rxs0005 wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


If one visualises this step by step:

Attachment:
STEP BY STEP.jpg
STEP BY STEP.jpg [ 12.52 KiB | Viewed 769 times ]


With the first step you just ask yourself how many different arrangements there are of 3 Letters? As bunuel calcualted this is simply 3! = 6
Then the constraints; put everything in so called "selection-boxes" and ask yourself, how many possible combinations does the first letter have, the second, and the last if A and E must be among the selected. Finally multiply with 6.
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How many different arrangements of letters are possible if [#permalink]

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reto wrote:
Could we also solve this with: Total Combinations - Forbidden Combinations?

Total = 5*4*3 = 60
Forbidden (A is not part): 4*3*2 = 24
Forbidden (B is not part): 4*3*2 = 24
Forbidden (A and B are not part): 3*2*1 = 6

Total Forbidden Combinations: 54, Answer 6

I know its wrong but where is my mistake?


Yes, you can do it this way. You are correct in all of your calculations, but you are double counting in your statements.

It should be like this:
Total = 5*4*3 = 60
Forbidden (A is not part of, but B is): 3*3*2 = 18
Forbidden (B is not part of, but A is): 3*3*2 = 18
Forbidden (Both A and B are not part of): 6

Total Forbidden Combinations = 42, Answer 6

You should be able to see where your problem is from this. =)
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Re: How many different arrangements of letters are possible if [#permalink]

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New post 20 Aug 2015, 18:08
reto wrote:
Bunuel wrote:
How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?

(A) 72
(B) 64
(C) 36
(D) 18
(E) 6


As A and E must be among 3 letters than the third letter must be out of B, C and D. 3C1=3 ways to choose which one it'll be. Now, 3 different letters can be arranged in 3!=6 ways, so final answer is 3*6=18.

Answer: D.


Could we also solve this with: Total Combinations - Forbidden Combinations?

Total = 5*4*3 = 60
Forbidden (A is not part): 4*3*2 = 24
Forbidden (B is not part): 4*3*2 = 24
Forbidden (A and B are not part): 3*2*1 = 6

Total Forbidden Combinations: 54, Answer 6

I know its wrong but where is my mistake?[/quote]

Hey there,

note that the formula from set theorey is Total - X - Y + [X AND Y].

you actually need to add 6 combinations back.
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Re: How many different arrangements of letters are possible if [#permalink]

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New post 07 Feb 2016, 09:51
I kind of got to the right answer differently...
we have:
A B C D E
5 letters.
we can thus select 3 out of 5 in: 5x4x3 ways. this is 60. Since the place of A and E is not important, we can divide by 2!, or 30 ways. Now, it must be true that we should have a number of combinations that is less than 30, because in 5x4x3 we have all combinations, including those in which A and E are not. so D looks fine.
Re: How many different arrangements of letters are possible if   [#permalink] 07 Feb 2016, 09:51
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