# How many different Braille characters can there be?

The Braille system for representing characters was developed in the early 19th century by Louis Braille. Special characters for the blind consist of raised dots. The positions for the dots are selected in two vertical columns of three dots each. There must be at least one raised dot. How many different Braille characters can there be?

In 1 you must look at the possibilities of the points being raised. Imagine you have a $$2 \times 3$$ grid that gives you the $$6$$ spaces. And you must divide your problem into $$6$$ disjoint sets, one represents the ways in which a point rises on the grid, set $$2$$ the ways in which two points rise on the grid and so on

I did this, but it's bad: \begin{align*} 3 \cdot 2 & = 6\\ 2^6 & = 64\\ \end{align*} $$64 - 1 = 63$$ characters. I have this, but now I need to find the conjunte 2 3 4

• You could start with all six spots being "equal", and then you'd have $2^6$ ways to turn them each "on/off." Then subtract all the ways that the first three could be all off ($2^3$), all the ways the second three could be all off (another $2^3$), and then add something back in that you counted twice. Sep 30, 2020 at 17:29
• Not sure the problem is clear. Surely any two configurations with exactly one raised dot would be the same, no? How could someone hope to distinguish two of these by feel alone? Which configurations are deemed to be equivalent?
– lulu
Sep 30, 2020 at 17:29
• @lulu That sounds right, but I checked, and the "a", comma, and apostrophe are all represented by a single dot. Of course, when you have a line of text, there's an implicit base line that allows you to distinguish the height of the text. I didn't notice any configurations that differ only because one is in the left column and one in the right. (I agree with your main point of course; we have to know what configurations are distinguishable.) Sep 30, 2020 at 17:47
• I put the conjunte 1 5,6 in the "photo2" but i need find the conjunte 2,3,4 Sep 30, 2020 at 17:49
• @pipe123xwe there are exactly $\binom 6k$ ways to choose $k$ dots from the $6$. Of course, that assumes that we are allowing all choices to count as different configurations.
– lulu
Sep 30, 2020 at 17:50

There are $$2^6$$ subsets of $$6$$ positions, but one of these is empty, so the count is $$2^6-1=63$$. Alternatively, you can condition on the number $$k$$ of dots, yielding $$\sum_{k=1}^6 \binom{6}{k}=6+15+20+15+6+1=63$$