Problem:
Call a 7-digit telephone number d1 d2 d3−d4 d5 d6 d7 memorable if the prefix sequence d1d2d3 is exactly the same as either of the sequences d4d5d6 or d5d6d7 (possibly both). Assuming that each di can be any of the ten decimal digits 0,1,2,…9, the number of different memorable telephone numbers is
Answer Choices:
A. 19810
B. 19910
C. 19990
D. 20000
E. 20100
Solution:
There are 10,000 ways to write the last four digits d4 d5 d6 d7, and among these there are 10000−10=9990 for which not all the digits are the same. For each of these, there are exactly two ways to adjoin the three digits d1d2d3 to obtain a memorable number. There are ten memorable numbers for which the last four digits are the same, for a total of 2⋅9990+10=19990.
OR
Let A denote the set of telephone numbers for which d1 d2 d3 and d4 d5 d6 are identical and B the set for which d1d2d3 is the same as d5d6d7. A number d1 d2 d3−d4 d5 d6 d7 belongs to A∩B if and only if d1=d4=d5=d2=d6=d3= d7. Hence, n(A∩B)=10. Thus, by the Inclusion-Exclusion Principle.
n(A∪B)=n(A)+n(B)−n(A∩B)=103⋅1⋅10+103⋅10⋅1−10=19990