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A random 10-letter code is to be formed using the letters A, B, C, D,

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A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 30 Oct 2014, 08:39
15
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

60% (01:43) correct 40% (01:45) wrong based on 311 sessions

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Tough and Tricky questions: Combinations.



A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the “I” will be used twice). What is the probability that a code that has the two I’s adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/2

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Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 30 Oct 2014, 16:53
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Bunuel wrote:

Tough and Tricky questions: Combinations.



A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the “I” will be used twice). What is the probability that a code that has the two I’s adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/5


Total number of combinations possible = \(10!/2!\)
Total favorable combinations = \(9!\) - We have considered the two Is as one element. Since we have two Is, they can interchange their positions, without affecting the code.

Probability = \(9!/(10!/2!) = 9!*2!/10! = 1/5\)
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Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 19 Jan 2015, 10:38
Thoughtosphere wrote:
Bunuel wrote:

Tough and Tricky questions: Combinations.



A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the “I” will be used twice). What is the probability that a code that has the two I’s adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/5


Total number of combinations possible = \(10!/2!\)
Total favorable combinations = \(9!\) - We have considered the two Is as one element. Since we have two Is, they can interchange their positions, without affecting the code.

Probability = \(9!/(10!/2!) = 9!*2!/10! = 1/5\)

did you calculate the number of favorable outcomes with 8!*9?

or how did you calculate them?
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Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 23 Jan 2016, 08:17
1
1
Bunuel wrote:

Tough and Tricky questions: Combinations.



A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the “I” will be used twice). What is the probability that a code that has the two I’s adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/2


we have 9 different letters and one of these is being used twice..

total ways code can be formed= 10!/2!..
10! because there are total 10 letters..
and div by 2! to cancel out repetitions, which are due to two 'I's...

take two Is together as one letter, remaining are 8 letters.. total 9.
ways =9!...

prob=9!/{10!/2!} = 9!2!/10!
2/10=1/5
C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 05 Jan 2017, 20:11
To find the probability of forming a code with two adjacent I’s, we must find the total number of such codes and divide by the total number of possible 10-letter codes.
The total number of possible 10-letter codes is equal to the total number of anagrams that can be formed using the letters ABCDEFGHII, that is 10!/2! (we divide by 2! to account for repetition of the I's).

To find the total number of 10 letter codes with two adjacent I’s, we can consider the two I’s as ONE LETTER. The reason for this is that for any given code with adjacent I’s, wherever one I is positioned, the other one must be positioned immediately next to it. For all intents and purposes, we can think of the 10 letter codes as having 9 letters (I-I is one). There are 9! ways to position 9 letters.

Probability = (# of adjacent I codes) / (# of total possible codes)
= 9! ÷ (10! / 2! ) = ( 9!2! / 10! ) = (9!2! / 10(9!) ) = 1/5
The correct answer is C.
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Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 10 Jul 2018, 07:19
10!/2! ways to arrange 10 letters <--- remember da formula
9! ways to arrange with "II" as 1 letter

9! * 2!/10! = 1/5
Answer C
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Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

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New post 23 Sep 2018, 04:21
Using anagrams, we can designate the letters as Not I = ni and I = i
so we have ni,ni,ni,ni,ni,ni,ni,ni, i , i apply 10C2 here which is 45
no we know that i and i are adjacent so we treat these two i's as one whole block

now this block of two i's can take nine positions
hence 9 different positions out of a total of 45 different positions
9/45 = 1/5
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Re: A random 10-letter code is to be formed using the letters A, B, C, D, &nbs [#permalink] 23 Sep 2018, 04:21
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