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08 Nov 2011, 01:39
Kudos
Bookmarks
E
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:
62% (01:55) correct
38%
(01:48)
wrong
based on 146
sessions
History
Date
Time
Result
Not Attempted Yet
What is the probability that a random arrangement of the letters IIITTNNOAV in a row results in the word INVITATION?
A. 1/4!
B. 1/6!
C. 1/10!
D. 4/10!
E. 4!/10!
A. 1/4!
B. 1/6!
C. 1/10!
D. 4/10!
E. 4!/10!
08 Nov 2011, 03:08
Kudos
Bookmarks
There are ten letters, so the possible number of arrangements is C=10!.
There is, however, more than one arrangement that results in the letter combination INVITATION:
There are 3! ways to arrange the I's and two ways for each of the letters N and T to be arranged. In total, this results in 3!2!2!=4!=24 letter combinations that will give you the word INVITATION.
Bringing this together, the probability is P = 24/10!.
There is, however, more than one arrangement that results in the letter combination INVITATION:
There are 3! ways to arrange the I's and two ways for each of the letters N and T to be arranged. In total, this results in 3!2!2!=4!=24 letter combinations that will give you the word INVITATION.
Bringing this together, the probability is P = 24/10!.
08 Nov 2011, 03:33
Kudos
Bookmarks
Sepideh2612
Denominator :
Arranging 10 letters IIITTNNOAV = 10! / 3!*2!*2!
numerator :
forming a word INVITATION = 1!
So its : 1!/ (10!/3!*2!*2!)
= 4!/10!
