Eight variables

Probability Level 2

Eight variables $a, b, c, d, e, f, g$ , and $h$ are assigned the integers from 1 to 8 in a random order. (No integer is assigned to more than one variable.)

If the probability that $a-b > 4$ is $\frac{x}{y},$ where $x$ and $y$ are coprime positive integers, what is $x+y?$

The answer is 31.

2 solutions

Geoff Pilling
Aug 1, 2017

There are six ways we can assign $a$ and $b$ such that $a-b > 4$ :

And there are $8\cdot 7 = 56$ ways to assign $a$ and $b$ .

So, the probability is $\dfrac{6}{56} = \dfrac{3}{28}$

$3+28 = \boxed{31}$

Excellent explanation of simple, elegant solution.

Holly Hart - 3 years, 9 months ago

the above solution is wrong there are only 28 ways to assign a and b so ans is (6/28) = (3/14) so x+y = 17

Er Amit Sharma - 3 years, 9 months ago

How do you get 28?

Geoff Pilling - 3 years, 8 months ago

Vallabh Deshpande
Sep 11, 2017

a > b +4

When b is 1, a can be 6, 7 or 8

when b is 2, a can be 7 or 8

and when b is 3, a can only be 8.

Also for each of these ways, others can be assigned values with 6! ways

Thus (3+2+1)*6! ways that are acceptable

Total number of ways to assign a value are 8!

therefore, (6*6!)/8! = 3/28

3 + 28 = 31

This solution, while correct, is needlessly complex. There is no reason to consider the remaining 6 assignments, as the question only involves a and b, which are distinct integers from [1, 8].

Lynn Kiaer - 3 years, 9 months ago

Yeah, I did not think of that while solving it!

Vallabh Deshpande - 3 years, 8 months ago

Eight variables

The answer is 31.

2 solutions

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