- #1
mr_coffee
- 1,629
- 1
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?
I'm thinking I have to use combinations becuase the multipcation rule won't work..the combination formula is the following:
Choosing r items out of n,
n!/r!(n-r)!
So I can choose 6 symbols, but if at least one has to be raised, that means all 6 can be raised, so would i have 12 possibilites?
then i was thinking maybe the total possilbites of r items would be
2^12 because the dot is either up or down.
So would the answer be 2^12 = 4096 symbols?
That seems way too big...
so the other answer might be:
12!/2!(10)! = 66. that sounds more like it but I'm still not sure if its correct.
Any help would be great
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?
I'm thinking I have to use combinations becuase the multipcation rule won't work..the combination formula is the following:
Choosing r items out of n,
n!/r!(n-r)!
So I can choose 6 symbols, but if at least one has to be raised, that means all 6 can be raised, so would i have 12 possibilites?
then i was thinking maybe the total possilbites of r items would be
2^12 because the dot is either up or down.
So would the answer be 2^12 = 4096 symbols?
That seems way too big...
so the other answer might be:
12!/2!(10)! = 66. that sounds more like it but I'm still not sure if its correct.
Any help would be great