Braille code probabilities homework

[roadrunner]

Hello everyone im' stuck on this problem.

It says:
Each symbol in braille code is represened by a rectangular arrangement of six dots. Given that a least 1 dot of the 6 must be raised, how many symbols can be represented in brail?

now i saw this posted somewhere else, they got 63. don't know how.

i got 63 two ways...

2^6-1 (case where all down)=63

6c1+6c2+6c3+6c4+6c5+6c6 -1 also =63 (where 6c3 etc is combinations...6 options choose 3)

can sumone explain how/why that works please.

also part b...how many combinations have EXACTLY 3 raised

and how many have an even number of raised dots


(this is for math and computer science course so not TOO sure which to post it in thanks)

 

[EnumaElish]

Think of "raised " as "included in a subset," and "flat" as "not included in a subset." Then "all flat" is the empty set. Number of subsets = 2^6 (including empty set).

How many combinations with exactly 3 raised = number of subsets with 3 elements.

How many with even raised = number of subsets with 2 or 4 or 6 elements.

 

[roadrunner]

so 2^6 is a? (-1 for original case where non are raised?)

then for b...2^3...so 8?
or do i need to figure out how many have more or less than 3 an subtract it?

 

[Dick]

The latter. Figure out how many have 0,1 and 2 raised and subtract the total from 2^6.

 

[HallsofIvy]

Another way of getting that answer is to argue that each of the 6 dots can be raised or not- a total of 6 "raised" or "not raised" choices. By the counting principal, there would be a total of 2⁶= 64 possiblities. That of course includes "none raised" which is not allowed so there are 64- 1= 63 allowed. That is the same as your sum of binomial coefficients because you are now arguing that it is "number with exactly one dot raised"+ "number with exactly two dots" raised"+ etc. "The number of 6 dots with exactly n raised" is the same as "how many ways can I choose n out of the 6 dots to raise": ₆C_n. And, of course, that is 2⁶- 1 because you did not include ₆C₁= 1. These are binomial coefficients- the coefficients in the expansion of (x+ y)⁶. The sum of the coefficients is that with x=y= 1: (1+1)⁶= 2⁶ (and again leaving out the first subtracts 1 from the whole thing).

As both Enuma Elish and Dick have said- to get "how many have at least three dots raised" Calculate ₆C₁ and ₆C₂ (which you may already have done it you calculated that sum of binomial coefficients directly) and subtract from the 63 you already had. (Dick included ₆C₀ and told you to subtract from 64. Same thing of course.)

 

[EnumaElish]

I have just realized that the OP is an exact replica of the OP in another thread in this forum, created simultaneously with this one.