A Tedious Inequality

Placing the digits 2 , 3 , 4 , 5 2, 3, 4, 5 exactly once in a \square , how many ways can we make the following inequality true:

× > × \square \times \square > \square \times \square

8 24 6 12

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1 solution

Kushal Bose
Oct 21, 2016

If four boxes are randomly filled up by four numbers there will be 4 ! = 24 4!=24 combinations.Among these combinations either leftside is greater than right side and vice-versa.

So above inequality holds in 24 / 2 = 12 24/2=12 cases.

Yup, we don't have to perform all the calculations.

However, we do have to make sure that one side is strictly greater than the other. If the numbers were 2 , 3 , 4 , 6 2, 3, 4, 6 instead, then the answer would not be 12. Try it !

Chung Kevin - 4 years, 7 months ago

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