Another tedious inequality

Probability Level 2

Placing the digits $2, 3, 4, 6$ exactly once in a $\square$ , how many ways can we make the following inequality true:

$\square \times \square > \square \times \square$

10 8 6 12

1 solution

Kushal Bose
Oct 21, 2016

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

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

Here sometimes equality will hold i.e. $2\times 6= 3 \times 4$ .They can permute in $2! \times 2!=4$ ways. Total ways $12-4=8$

That's the idea! Can you add in more details so that others who can't solve it can understand what you did?

Chung Kevin - 4 years, 7 months ago

I have updated my solution

Kushal Bose - 4 years, 7 months ago

We have to be careful about the strict inequality in the phrasing. I think it's confusing to say "inequality holds in 12 cases" but "total ways = 8".

Also, there are $2 \times 2! \times 2! = 8$ ways to permute the $2 \times 6 = 3 \times 4$ , because we can flip the LHS and RHS.

I think it is better to start off by explaining why there are 24 - 8 combinations of strict inequalities, and then dividing by 2.

Chung Kevin - 4 years, 7 months ago

@Chung Kevin – Yes I have missed out fliping

Kushal Bose - 4 years, 7 months ago

Another tedious inequality

1 solution

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