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24 Nov 2021, 08:07
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In how many ways can the letters of the word PERMUTATIONS be arranged such that the order of vowels remains unchanged?
A. \(\frac{12!}{2!}\)
B. \(\frac{12!}{5!}\)
C. \(\frac{12!}{2!*5!}\)
D. \(\frac{4*7!}{2!}\)
E. \(\frac{4*7!}{2!*5!}\)
A. \(\frac{12!}{2!}\)
B. \(\frac{12!}{5!}\)
C. \(\frac{12!}{2!*5!}\)
D. \(\frac{4*7!}{2!}\)
E. \(\frac{4*7!}{2!*5!}\)
24 Nov 2021, 08:07
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Official Solution:
In how many ways can the letters of the word PERMUTATIONS be arranged such that the order of vowels remains unchanged?
A. \(\frac{12!}{2!}\)
B. \(\frac{12!}{5!}\)
C. \(\frac{12!}{2!*5!}\)
D. \(\frac{4*7!}{2!}\)
E. \(\frac{4*7!}{2!*5!}\)
PERMUTATIONS has 12 letters: 5 vowels (EUAIO) and 7 consonants (PRMTTNS).
Choose 5 slots from 12 for the vowels: \(C^5_{12}=\frac{12!}{5!*7!}\). Since the order of the vowels must remain unchanged (EUAIO), then we do not arrange them.
7 consonants in 7 remaining slots can be arranged in \(\frac{7!}{2!}\) ways (notice that we have two T's among 7 consonants).
So, the total number of ways is \(\frac{12!}{5!*7!}*\frac{7!}{2!}=\frac{12!}{2!*5!}\).
Answer: C
In how many ways can the letters of the word PERMUTATIONS be arranged such that the order of vowels remains unchanged?
A. \(\frac{12!}{2!}\)
B. \(\frac{12!}{5!}\)
C. \(\frac{12!}{2!*5!}\)
D. \(\frac{4*7!}{2!}\)
E. \(\frac{4*7!}{2!*5!}\)
PERMUTATIONS has 12 letters: 5 vowels (EUAIO) and 7 consonants (PRMTTNS).
Choose 5 slots from 12 for the vowels: \(C^5_{12}=\frac{12!}{5!*7!}\). Since the order of the vowels must remain unchanged (EUAIO), then we do not arrange them.
7 consonants in 7 remaining slots can be arranged in \(\frac{7!}{2!}\) ways (notice that we have two T's among 7 consonants).
So, the total number of ways is \(\frac{12!}{5!*7!}*\frac{7!}{2!}=\frac{12!}{2!*5!}\).
Answer: C
17 Feb 2025, 22:26
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Another way to think could be,
Total possible arrangements = 12!/2! (T is repeated twice)
Now notice that there are 5 different vowels and we want only one order out of the default 5! orders, so we will need to remove rest of the orders but one, which is nothing but,
12!/2! * 1/5! = 12!/(2!*5!)
It's like saying for ABC, out of all the orders possible, I just need ABC ie. 1 out of 3!
Ans: C
Total possible arrangements = 12!/2! (T is repeated twice)
Now notice that there are 5 different vowels and we want only one order out of the default 5! orders, so we will need to remove rest of the orders but one, which is nothing but,
12!/2! * 1/5! = 12!/(2!*5!)
It's like saying for ABC, out of all the orders possible, I just need ABC ie. 1 out of 3!
Ans: C
