Question 457646
This is a peculiar question because the number of choices does not matter.  You either get the answer right or wrong.  60% of 5 = 3/5, so you need to get 3 of the 5 questions right.
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You can get them Right or Wrong in several ways, but do you need to know that, too?
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RRRWW
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Shift left and wrap the one that went of the left end onto the right end
RRWWR
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Repeat
RWWRR
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Repeat
WRRRW
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If you repeat again, you're right back to where you were, so that is not a "new" sequence.
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Or you could start with RWRWR and go through those
RWRWR
WRWRR
RWRRW
WRRWR
RRWRW
RWRWR is a repeat sequence
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Or you start with: WRWRR, or WWRRR, etc.
But that is a lot of work, and error-prone, too.
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If you have done anything with permutations and combinations, you may recognize that the question is asking how many ways you can pick 3 specific items from a set of 5 items:  C(n,r), where n=5, r=3
{{{C(n,r) = n!/((r!)*(n-r)!)}}}
{{{C(5,3) = 5!/(3!*2!)}}}
{{{C(5,3) = (5*4*3*2*1)/((3*2*1)*(2*1))}}}
Ignoring the "*1" factors
{{{C(5,3) = (5*4*3*2)/(3*2*2)}}}
{{{C(5,3) = (5*2)}}}
{{{C(5,3) = 10}}} 
That means there are 10 ways you can get 3 questions right out of 5 total questions.
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Now if the question were to have asked "at least 60%" then the answer would be different because you would have to consider the case of getting all 5 right as well as 4/5 right.
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Done