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05 Dec 2022, 07:28
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (01:43) correct
39%
(01:49)
wrong
based on 180
sessions
History
Date
Time
Result
Not Attempted Yet
A certain computer program produces anagrams by taking words and randomly rearranging their letters. If "SKATE" is entered into the program, what is the probability that any resulting anagram will not contain adjacent vowels ?
(A) 1/5
(B) 2/5
(C) 1/2
(D) 3/5
(E) 4/5
(A) 1/5
(B) 2/5
(C) 1/2
(D) 3/5
(E) 4/5
13 Dec 2022, 03:26
Kudos
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The probability will be: Total ways in which no vowels will be adjacent/Total possible outcomes
Total possible outcomes: \(5!\)
Total ways in which no vowels will be adjacent:
The fastest way to find this is to subtract the number many ways the vowels can be adjacent from the total number of possible outcomes. This will leave you with the number of the ways in which the vowels will not be adjacent.
To do this, group the vowels together as if they were a single letter. Now there are \(4!\) possible outcomes. However as tAEks is different to tEAks, one needs to account for this by multiplying by \(2\).
\(5!-4!*2\)
\(= 4!(5-1*2)\)
\(= 4!(3)\)
Probability:
\(\frac{4!(3)}{5!}\)
\(= \frac{3}{5}\)
Answer D
Total possible outcomes: \(5!\)
Total ways in which no vowels will be adjacent:
The fastest way to find this is to subtract the number many ways the vowels can be adjacent from the total number of possible outcomes. This will leave you with the number of the ways in which the vowels will not be adjacent.
To do this, group the vowels together as if they were a single letter. Now there are \(4!\) possible outcomes. However as tAEks is different to tEAks, one needs to account for this by multiplying by \(2\).
\(5!-4!*2\)
\(= 4!(5-1*2)\)
\(= 4!(3)\)
Probability:
\(\frac{4!(3)}{5!}\)
\(= \frac{3}{5}\)
Answer D
01 Feb 2025, 09:29
Kudos
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Slightly different approach:
1) Vowels = A & E
2) Ways to order 5 letters: 5! = 120
3) Ways to order 3 consonants (K, S & T): 3! = 6
4) Spaces in between ordered K, S & T: _K_S_T_ --> 4 spaces
5) Number of ways the two letters A & E can be split over the 4 spaces: 4C2 * 2 = 12
(Note: factor of two is there because AKSTE is different from EKSTA)
6) Probability = (3!*4C2*2)/5! = 3/5
A similar question with a twist is this one: https://gmatclub.com/forum/in-how-many- ... olG8UbogsF
There one has to spread out 4 equal I's over 8 spots, resulting in a factor of 8C2, without a factor of 2 as above.
1) Vowels = A & E
2) Ways to order 5 letters: 5! = 120
3) Ways to order 3 consonants (K, S & T): 3! = 6
4) Spaces in between ordered K, S & T: _K_S_T_ --> 4 spaces
5) Number of ways the two letters A & E can be split over the 4 spaces: 4C2 * 2 = 12
(Note: factor of two is there because AKSTE is different from EKSTA)
6) Probability = (3!*4C2*2)/5! = 3/5
A similar question with a twist is this one: https://gmatclub.com/forum/in-how-many- ... olG8UbogsF
There one has to spread out 4 equal I's over 8 spots, resulting in a factor of 8C2, without a factor of 2 as above.
