# 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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