1983 AIME Problems/Problem 10
Problem
The numbers 
, 
and 
have something in common: each is a 
-digit number beginning with 
that has exactly two identical digits. How many such numbers are there?
Solution
Solution 1
Suppose that the two identical digits are both . Since the thousands digit must be , only one of the other three digits can be . This means the possible forms for the number are
Because the number must have exactly two identical digits, 
, 
, and 
. Hence, there are 
numbers of this form.
Now suppose that the two identical digits are not . Reasoning similarly to before, we have the following possibilities:
Again, , 
, and 
. There are numbers of this form.
Thus the answer is 
.
Solution 2
Consider a sequence of digits instead of a -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, from which 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 | ||
