1983 AIME Problems/Problem 10
Problem
The numbers 
, 
, and 
have something in common. Each is a four-digit number beginning with 
that has exactly two identical digits. How many such numbers are there?
Solution
Solution 1
Suppose the two identical digits are both one. Since the thousands digits must be one, the other one can be in only one of three digits,
Because the number must have exactly two identical digits, 
, 
, and 
. Hence, there are 
numbers of this form.
Suppose the two identical digits are not one. Therefore, consider the following possibilities,
Again, , 
, and 
. There are numbers of this form.
Thus, the desired answer is 
.
Solution 2
Consider a sequence of four digits instead of a four digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is 
. This means we can find all possible sequences with one digit repeated twice, and then divide by 
.
If we let the three distinct digits of the sequence be 
and 
with 
repeated twice, we can make a table with all possible sequences:
![\[\begin{tabular}{ccc} aabc & abac & abca \\ baac & baca & \\ bcaa && \\ \end{tabular}\]](http://latex.artofproblemsolving.com/1/3/f/13f119994b5b039f67296f2a0b00f107bf20776e.png)
There are 
possible sequences.
Next, we can see how many ways we can pick and 
. This is 
because there are digits, and we need to choose 
with regard to order. This means there are 
sequences of length 
with one digit repeated. We divide by 10 to get 
as our answer.
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
