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A new word is to be formed by randomly rearranging the lett
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14 May 2019, 00:28
14 May 2019, 00:28
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Expert Reply
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Question Stats:
74% (02:44) correct
25% (02:14) wrong based on 47 sessions
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A new word is to be formed by randomly rearranging the letters of the word ALGEBRA. What is the probability that the new word has consonants occupying only the positions currently occupied by consonants in the word ALGEBRA?
(A) \(\frac{2}{120}\)
(B) \(\frac{1}{24}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{105}\)
(E) \(\frac{1}{35}\)
(A) \(\frac{2}{120}\)
(B) \(\frac{1}{24}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{105}\)
(E) \(\frac{1}{35}\)
Re: A new word is to be formed by randomly rearranging the lett
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19 Oct 2019, 01:58
19 Oct 2019, 01:58
1
We have 7 letters where A is repeated. Hence we can arrange them in 7!/2! ways.
The condition we have is that the consonants are arranged in the same four places.
Therfore, we arrnage those 4 letters in 4! ways and we arrange the other 3 in 3!/2! = 3.
Summing up then, 4!*3/7!/2! = 4! * 3 * 2! / 7! = 6/(7*6*5) = 1/35
Answer is E.
The condition we have is that the consonants are arranged in the same four places.
Therfore, we arrnage those 4 letters in 4! ways and we arrange the other 3 in 3!/2! = 3.
Summing up then, 4!*3/7!/2! = 4! * 3 * 2! / 7! = 6/(7*6*5) = 1/35
Answer is E.
Re: A new word is to be formed by randomly rearranging the lett
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10 Feb 2020, 18:42
10 Feb 2020, 18:42
4
Carcass wrote:
A new word is to be formed by randomly rearranging the letters of the word ALGEBRA. What is the probability that the new word has consonants occupying only the positions currently occupied by consonants in the word ALGEBRA?
(A) \(\frac{2}{120}\)
(B) \(\frac{1}{24}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{105}\)
(E) \(\frac{1}{35}\)
(A) \(\frac{2}{120}\)
(B) \(\frac{1}{24}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{105}\)
(E) \(\frac{1}{35}\)
Total number of ways of arranging the word ALGEBRA is \(\frac{7!}{2!}\)
Now the consonants occupy the same position as in the word ALGEBRA. So consonants will be _LG_BR_.
The 3 vowels in the 3 blanks can be arranged in \(\frac{3!}{2!}\) ways.
The consonants can be arranged in those positions in 4!
Probability = \(\frac{(\frac{3!}{2!}) * 4!}{\frac{7!}{2!}}\)
= \(\frac{3*4!}{3*4*5*6*7}\)
= \(\frac{1}{35}\) (E)
