2006 AIME II Problem 4

Below is the professionally curated solution for Problem 4 of the 2006 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME II solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:arrangements with restrictionscombinations

Difficulty rating: 2180

4.

Let (a1,a2,a3,,a12)(a_1, a_2, a_3, \ldots, a_{12}) be a permutation of (1,2,3,,12)(1, 2, 3, \ldots, 12) for which a1>a2>a3>a4>a5>a6anda6<a7<a8<a9<a10<a11<a12.a_1 \gt a_2 \gt a_3 \gt a_4 \gt a_5 \gt a_6 \quad \text{and} \quad a_6 \lt a_7 \lt a_8 \lt a_9 \lt a_{10} \lt a_{11} \lt a_{12}. An example of such a permutation is (6,5,4,3,2,1,7,8,9,10,11,12).(6, 5, 4, 3, 2, 1, 7, 8, 9, 10, 11, 12). Find the number of such permutations.

Solution:

The term a6a_6 is smaller than every other term of the permutation, so a6=1.a_6 = 1. Now choose which five of the remaining 1111 numbers occupy positions 11 through 5:5: they must appear in decreasing order, so their arrangement is forced, and the other six numbers must fill positions 77 through 1212 in increasing order, which is also forced.

Every choice of the five numbers gives exactly one valid permutation, so the count is (115)=462.\binom{11}{5} = 462.

← Problem 3Full ExamProblem 5

Problem 4 in Other Years