Combinations and permutations

[dbag]

Hello

Problem statement:
group of 8 numbers (1 or 0)
a. How many groups are there
b. How many groups, such that there are 2 0's and 4 1's
c. atleast 6 0's?

Attempt at solving:
a. 8 numbers 2 possibilities so 2^8 = 256
b. no idea
c. 8 numbers 2 possibilites for last 2 so 1^6 * 2^2 = 4

any tips for the b part?

 

[pka]

Problem statement:
group of 8 numbers (1 or 0)
a. How many groups are there
b. How many groups, such that there are 2 0's and 4 1's
c. atleast 6 0's?

Attempt at solving:
a. 8 numbers 2 possibilities so 2^8 = 256
b. no idea
c. 8 numbers 2 possibilites for first 6 so 1^6 * 2^2 = 4

This one of the oddest stated questions I have ever seen.

I cannot see that combinations or permutations have anything to do with this.
If this were a question about about 8-bitstrings (strings of 0's or 1's of length eight) then we could use those concepts.

The way these are written I could make a strong argument that the answers are:
a) nine, b) three, & c) three.
Please come back tell us how you were expected to answer.

 

[dbag]

If this were a question about about 8-bitstrings (strings of 0's or 1's of length eight) then we could use those concepts
that's what its about

a) how many different types of arrangements are there? for example 00001111, 10101010
b) How many arrangements such that 2 of the 8 are zeroes and 6 are ones for example 00111111, 10011111
c) how many arrangements such that there are 6 zeroes and 2 anything else 00000011, 00000010

 

[pka]

If this were a question about about 8-bit-strings (strings of 0's or 1's of length eight) then we could use those concepts
that's what its about
a) how many different types of arrangements are there? for example 00001111, 10101010
b) How many arrangements such that 2 of the 8 are zeroes and 6 are ones for example 00111111, 10011111
c) how many arrangements such that there are 6 zeroes and 2 anything else 00000011, 00000010

a) There are \(\displaystyle 2^8=256\) different bit-strings of length eight.
b) there are \(\displaystyle \frac{8!}{2!\cdot 6!}\) different 8-bit-strings consisting of two 0's & six ones.
c) there are \(\displaystyle \dbinom{8}{6}+\dbinom{8}{7}+\dbinom{8}{8}\) different 8-bit-strings consisting of six 0's & two of any other.