2022 AMC 12A Problem 10

Question

What is the number of ways the numbers from

1

to

14

can be split into

7

pairs such that for each pair, the greater number is at least

2

times the smaller number?

108

120

126

132

144

A number

m

can be the smaller of its pair only if

2m\le14

So

1

–

7

are the smaller elements and

8

–

14

the larger; assign greedily from the tightest smaller number

Any number

8

or larger cannot be a smaller element (its double exceeds

14

), so

8

–

14

are all larger elements and

1

–

7

are all smaller elements.

Match each smaller

s

to a larger

g\ge2s.

Processing from the most restrictive:

s=7

forces

g=14

(

1

way); then

s=6

has

\{12,13\}

left (

2

);

s=5

has

3;

s=4

has

4;

s=3

has

3;

s=2

has

2;

s=1

has

1.

The number of matchings is

1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144.

Thus, the correct answer is E.

Problem 10 in other years