In how many ways can the numbers 1 through 5 be entered once each into the five boxes below so that all the given inequalities are true?

Guest Jul 10, 2023

#1**0 **

I presume the boxes and inequalities look like such

[ ] < [ ] > [ ] < [ ] > [ ]

We can set the inequality as

a < b > c < d > e

To solve the question, there are 2 cases:

case 1:

b=4 d=5 (or the other way around)

Then we can set a, c, and e, as any number (1, 2 or 3)

so for case 1 we have 2*3!=2*6=12

case 2:

b=5, a=4 (or on the other side, d=5, e=4)

In this case we know that d has to be 3, other wise the inequality won't *work*

we have two ways to put 1, and 2 into the equation, so

For case 2, there are 2*2=4 ways

Add them together: 4+12=16

(wow this sounded a lot simpler in my head) Please confirm this

ANITBIOTICS Jul 10, 2023