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08 Nov 2011, 01:39
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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58% (01:46) correct
42%
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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
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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
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Sepideh2612
What is the probability that a random arrangement of the letters IIITTNNOAV in a row results in the word INVITATION?
The original answer is 4!/10!
I don´t have any explanation to this and I´d appreciate any help!
The original answer is 4!/10!
I don´t have any explanation to this and I´d appreciate any help!
Denominator :
Arranging 10 letters IIITTNNOAV = 10! / 3!*2!*2!
numerator :
forming a word INVITATION = 1!
So its : 1!/ (10!/3!*2!*2!)
= 4!/10!
