Problem:
How many of the numbers, 100,101,…,999, have three different digits in increasing order or in decreasing order?
Answer Choices:
A. 120
B. 168
C. 204
D. 216
E. 240
Solution:
For every 3 distinct digits selected from {1,2,…,9} there is exactly one way to arrange them into a number with increasing digits, and every number with increasing digits corresponds to one of these selections. Similarly, the numbers with decreasing digits correspond to the subsets with 3 elements of the set of all 10 digits. Hence our answer is
and grouping them according to their first digit, we see there are 1+(1+2)+(1+2+3)+⋯+(1+2+⋯+7)=84 such numbers. Now make a list of such numbers with decreasing digits in increasing order:
and group them according to their first digit. In this case there are 1+3+6+10+⋯+28+36=120 such numbers, an additional 36 numbers since a number with decreasing digits can contain 0 while one with increasing digits cannot. Thus, the answer is 84+120=204.