GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 May 2019, 21:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55231
A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

30 Oct 2014, 09:39
16
00:00

Difficulty:

55% (hard)

Question Stats:

62% (01:45) correct 38% (01:48) wrong based on 327 sessions

### HideShow timer Statistics

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

_________________
Manager
Joined: 02 Jul 2012
Posts: 186
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)
Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

30 Oct 2014, 17:53
7
2
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 = $$9!/(10!/2!) = 9!*2!/10! = 1/5$$
_________________
Give KUDOS if the post helps you...
##### General Discussion
Manager
Joined: 17 Dec 2013
Posts: 56
GMAT Date: 01-08-2015
Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

19 Jan 2015, 11:38
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?
Math Expert
Joined: 02 Aug 2009
Posts: 7684
Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

23 Jan 2016, 09:17
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
_________________
Director
Joined: 26 Oct 2016
Posts: 633
Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

05 Jan 2017, 21:11
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
_________________
Thanks & Regards,
Anaira Mitch
Manager
Joined: 07 Feb 2017
Posts: 183
Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

10 Jul 2018, 08:19
10!/2! ways to arrange 10 letters <--- remember da formula
9! ways to arrange with "II" as 1 letter

9! * 2!/10! = 1/5
Intern
Joined: 14 Jun 2018
Posts: 7
Re: A random 10-letter code is to be formed using the letters A, B, C, D,  [#permalink]

### Show Tags

23 Sep 2018, 05:21
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
Re: A random 10-letter code is to be formed using the letters A, B, C, D,   [#permalink] 23 Sep 2018, 05:21
Display posts from previous: Sort by