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

Question

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

Show Answer

Official Answer

C

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

Thoughtosphere · Accepted Answer

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

KarishmaB · Answer

No, numerator will be 9! only.

Think of it this way: In how many ways can you arrange 10 crayons of different colours such that C2 (Red) and C3 (Blue) must be placed together?

You tie C2 and C3 together and then arrange 9 groups/individuals. You do it in 9! ways. 
Then you multiply by 2 to arrange C2 and C3 as C3 and C2 too. But what if C2 and C3 are both identical, say both Red? Then will you multiply by 2? No.

This is the same case. You have 9 distinct letters to be arranged and you can do it in 9! ways. Wherever you write I, just write it once again. No extra arrangements take place.

eddyki · Answer

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

or how did you calculate them?

chetan2u · Answer

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

anairamitch1804 · Answer

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.

gmatzpractice · Answer

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

aadilismail · Answer

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

Nikhil30 · Answer

hey ,
P=no. of fav outcome/total no. of outcome.
for no. of fav outcome why is it not 9!/2!??

andreagonzalez2k · Answer

Name II duo X (they are always together in favorable outcomes).
Favorable outcomes are the permutations of A, B, C, D, E, F, G, H, X = 9!.

kittle · Answer

Hey  KarishmaB 
Why are we not doing: (9! / 2!) / (10!/2!) ? That way the answer should be A.
In the arrangement with 2 I's together for the numberator part (i.e the favourable outcomes), we would do the division by 2! right?

bumpbot · Answer

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

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