# What are we learning ?

**Competency in Focus:**Basic counting principle and permutation This problem from American Mathematics contest (AMC 8, 2012)

# First look at the knowledge graph.

# Next understand the problem

How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?

##### Source of the problem

American Mathematical Contest 2012, AMC 8 Problem 10

##### Key Competency

Permutation and basic counting principle

##### Difficulty Level

4/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

# Start with hints

Do you really need a hint? Try it first!

For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of , since all of the valid 4-digit number will always be greater than .

The best way to solve this problem is by using casework. Now think what are the cases?

It has two cases , as there can be only two leading digits, namely or .

CASE 1: As 2012 consits of two 2’s , one 1, 0 so if we set 1 as the leading digit then we have two twos and one 0 such numbers then we have such numbers.

When the leading digit is then we have one 2, one 1 and one 0 then we can arrange them in ways and as such we have 6 such numbers.

By addition principle we find that there are 9 such numbers.

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## AMC - AIME Program

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