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:
We split the set of numbers of the type described into two subsets and make two separate subcounts:
Those numbers having two 's. The second may be placed in any one of three positions and the two other numbers (distinct) may be placed in ways. Thus, there are such numbers.
Those numbers having two of some digit other than . The pair of identical digits may be selected in ways and then p1aced in ways (positions and , and , or and ). The remaining number may be selected in ways. Thus there are such numbers.
Thus, the answer is . (It is coincidental, perhaps remarkable, that the two subtotals are the same. This is not so for, say, the -digit version of this problem. Try it!)
Consider cells (for the -digit number) as below:
Select two of the cells as the ones that are to receive the equal digits. Call the left of these cells and the right one . This can be done in ways. For instance, one might have
Starting with the first cell, fill in all cells except R with distinct numbers. This can be done in ways (since there is only one choice, the number , for the first cell, which is never ). Finally, fill with the same number as was inserted in . This can be done in just way. Thus, the "product of ways" is .
The problems on this page are the property of the MAA's American Mathematics Competitions