# An Alphabet contains only A and B. Can \$ABB = BAA\$?

by Angelo Mark   Last Updated August 14, 2019 06:20 AM

An Alphabet containing only $$A$$ and $$B$$.

Removing $$AB$$ makes no difference to a word and adding $$BA$$ or $$AABB$$ makes no difference to a word.

Is $$ABB = BAA$$?

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ABB can only be equal to words that have one more B than A. So BAA is not achievable.

Matthew Daly
August 14, 2019 05:40 AM

Observe that if two words have no difference, then the difference of the number of letter $$A$$ in the word and the number of letter $$B$$ must be same. In the problem, $$ABB$$ has the difference of -1 but $$BAA$$ is 1. Therefore, $$ABB$$ is not equal to $$BAA$$.

Isaac YIU Math Studio
August 14, 2019 06:07 AM

Am I Correct?

$$ABA=A$$ (By removing $$AB$$)

Now by adding $$BA$$ will not make any difference, so

BA$$ABA=A$$

$$BA$$AB$$A=A$$

By removing $$AB$$ we get $$BAA=A$$(1)

Clearly , $$ABB=B$$ (2)(By removing $$AB$$)

Now by (1) and (2), we have $$BAA=A$$ and $$ABB=B$$.

Since $$A$$ and $$B$$ are unique, $$BAA \neq ABB$$

Angelo Mark
August 14, 2019 06:15 AM

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