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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
A random 10-letter code is to be formed using the letters A, B, C, D,
[#permalink]
04 May 2021, 03:23
04 May 2021, 03:23
Expert Reply
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Question Stats:
45% (01:28) correct
54% (01:12) wrong based on 11 sessions
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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
(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/2
A random 10-letter code is to be formed using the letters A, B, C, D,
[#permalink]
04 May 2021, 07:23
04 May 2021, 07:23
1
Solution:
We can consider the 2 I's as one, to find that the total number of ways the letters could be arranged is 9! i.e the first place could be select in 9 ways by the 9 letters the second in 8 and so on and so forth.
Next, the total number of ways the letters can be arranged= 10C2=\(\frac{10!}{2}\)=5*9!
Thus, the probability= \(\frac{9!*5}{9!}\)=\(\frac{1}{5}\)
IMO C
We can consider the 2 I's as one, to find that the total number of ways the letters could be arranged is 9! i.e the first place could be select in 9 ways by the 9 letters the second in 8 and so on and so forth.
Next, the total number of ways the letters can be arranged= 10C2=\(\frac{10!}{2}\)=5*9!
Thus, the probability= \(\frac{9!*5}{9!}\)=\(\frac{1}{5}\)
IMO C
gmatclubot
A random 10-letter code is to be formed using the letters A, B, C, D, [#permalink]
04 May 2021, 07:23
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