Solution to an introductory problem in combinatorics

This refers to the problems posed in the previous blog HW:

  1. Let A denote the set for which the set d_{1}d_{2}d_{3} is the same as d_{4}d_{5}d_{6} and B denote the set for which d_{1}d_{2}d_{3} is the same as d_{5}d_{6}d_{7}. A telephone number belongs to the set A \bigcap B if and only if all the seven d’s are equal, which can happen in 10 ways only. Hence, n(A \bigcap B)=10. Thus, by the Inclusion-Exclusion Principle, n(A\bigcup B) = n(A)+n(B)-n(A \bigcap B)= 10^{3}.10.1 +10^{3}.10.1 - 10=19990.
  2. I will post this solution in a few days.

Regards,

Nalin Pithwa

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