AMC-8 USA Math Olympiad

Permutation and basic counting principle AMC 8 2012, problem 10

Try this beautiful problem from AMC 8. It involves permutation and basic counting principles. We provide sequential hints so that you can try the problem.

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
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 $2012$, since all of the valid 4-digit number will always be greater than $1000$.  
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 $1$ or $2$.
We know  that number of ways of arranging ‘n’ items, out of which ‘p’ are alike, ‘q’ are alike and ‘r’ are alike given that p + q + r = n Number of ways of distributing ‘n’ distinct items, in groups of size ‘p’, ‘q’ and ‘r’ given that p + q + r = n 5. . Now try to calculate the the two cases .
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  $\frac{3!}{2!1!} \implies 3$ such numbers.  
When the leading digit is $2$ then we have one 2, one 1 and one 0 then we can arrange them in  $3! \implies 6$ ways and as such we have 6 such numbers.
By addition principle we find that there are  9 such numbers.

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