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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] New post 27 Nov 2010, 06:49
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Question Stats:

61% (02:08) correct 39% (01:21) wrong based on 151 sessions
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] New post 27 Nov 2010, 06:57
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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] New post 03 Dec 2010, 03:01
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] New post 03 Dec 2010, 04:10
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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] New post 30 Jun 2013, 23:59
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: How many different arrangements of letters are possible if [#permalink] New post 04 Dec 2014, 11:43
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Re: How many different arrangements of letters are possible if [#permalink] New post 13 Jul 2015, 10:51
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] New post 20 Jul 2015, 21:22
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] New post 21 Jul 2015, 22:04
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] 21 Jul 2015, 22:04
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