Question

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?

$108$

$120$

$126$

$132$

$144$

Solution:

We can see that the numbers $1-8$ must be in different pairs. $7$ must also be paired with $14$ since no other number is at least twice $7.$

Now let's look at what the other numbers can pair with. $8$ and $9$ can pair with any number $1-4.$ $10$ and $11$ can pair with any number $1-5,$ and $12$ and $13$ can pair with any number $1-6.$

$8$ can pair with $4$ numbers, but then $9$ only has $3$ options since $8$ took one. $10$ then has $3$ options, since $2$ choices are taken, but it has one more to choose from $(5).$ $11$ then has $2$ options, $12$ has $2$ options, and $13$ only has $1.$

Multiplying these together yields $4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$

Thus, E is the correct answer.

2022 AMC 10A Problem 14

Solution:

Problem 14 in Other Years