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
Permutation and basic counting principle
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.