2024 AMC 12A Problem 16

Question

A set of

12

tokens —

3

red,

2

white,

1

blue, and

6

black — is to be distributed at random to

3

game players,

4

tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as

\dfrac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m+n?

387

388

389

390

391

Answer: C

Concepts:basic probability multiset permutations

Difficulty rating: 1820

Small hint:

Assign the three roles (all reds, all whites, the lone blue) to the players in

3!

ways

Big hint:

The black tokens must then split as

1,2,3

among the players; divide by

\binom{12}{4,4,4}

Solution:

Treat all tokens as distinct; the total number of ways to deal

4

to each player is

\binom{12}{4,4,4}=34650.

For the favorable event, choose which player gets the reds, whites, and blue in

3!=6

ways. The red player needs

1

more token, the white player

2

more, and the blue player

3

more, all black; the

6

black tokens split as

1,2,3

in

\tfrac{6!}{1!\,2!\,3!}=60

ways. So the probability is

\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}.

Then

m+n=4+385=389.

Thus, the correct answer is C.