How many possibilities are there for the 3?

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SUMMARY

The discussion revolves around calculating the probability of arranging six digits (2, 3, 4, 5, 6, 7, 8) such that the digits 2 and 3 appear in the correct order but are not consecutive. The denominator for the probability calculation is established as 7!, representing all possible arrangements. The numerator is debated, with participants suggesting combinations like 5C4 and exploring various placements for the digit 3 relative to the digit 2. The final probability is contested, with figures of 5/21 and 5/14 mentioned as potential answers.

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Homework Statement


Six digits from the numbers 2, 3, 4, 5, 6, 7, 8 are chosen and arranged in a row without replacement. Find the probability that the digits 2 and 3 appear in the proper order but not consecutively


Homework Equations





The Attempt at a Solution


i know that the denominator is 7!, however I'm having trouble finding the numerator. I think that the numerator is 5C4 * 1 *1 however the answer is 5/21
 
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Suppose that 2 is in the 1st place. How many possibilities are there for the 3?
If 2 is in the 2nd place, in how many places can you put 3?
And if 2 is in the 3rd position?
etc.
 


i believe the answer is actually 5/14
 


I got something else than 5/21 as well (5/14 sounds familiar). But we will know as soon as Jennifer posts again.

Perhaps she has misquoted the answer, or didn't state the complete question.
 

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