# Combinatorics - Sequences with repetition and restrictions

Question: How many sequences of five elements with repetition allowed can be created from elements of the set $$\{1,2,3,4,5,6\}$$ in which the last digit is equal to any of the previous digits?

My Answer: Let's use the inclusion-exclusion principle. First we need to know the number of all sequences, that is: $$6^5$$ since we're allowing repetition. Then let's define the sets for $$k\in\{1,2,3,4,5,6\}$$: $$C_k = \{ \text{sequences with last digit equal k and previous digits different than k}\}$$ Clearly the sets $$C_k$$ are disjoint, since there is no sequence equal to another one with the last digits being different. Therefore, the answer is: $$6^5 - |C_1\cup C_2 \cup C_3 \cup C_4 \cup C_5 \cup C_6| = 6^5 - 6\cdot 5^4$$

Is my answer correct? Any help is highly appreciated! Thanks!

• Yes, this is perfect. – Don Thousand Oct 20 '18 at 1:05
• @Rushabh Mehta thank you! – Bruno Reis Oct 20 '18 at 2:50

Yes this looks correct. Another method is to directly count the sequences where the last digit is equal to any other:

$$6 - - - 6 \to 6^3$$

$$-\ 6 - -\ 6 \to 5\cdot 6^2$$

$$- - 6 - 6 \to 5^2\cdot 6$$

$$- - -\ 6\ 6 \to 5^3$$

There are $$6$$ of the above sequences, one for each number in the set.

$$6(6^3 + 5\cdot 6^2 + 5^2\cdot 6 + 5^3) = 4026$$

• Thats the direct method of doing it! It's a nice way too... Thanks for sharing it. – Bruno Reis Oct 20 '18 at 2:49