How many bit strings of length $10$ have exactly six 0’s?

Question

The answer to this question is $10 \choose 6$. However, I find this odd. Why would it be a combination in this case and not a permutation?

Ordering does not matter in combinations from my understanding, this would mean that $001 = 010$, which seems counterintuitive.

I do not understand why I have to use combinations for this problem.

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To clarify the issue, when I have two people {Bob, Alice}. I can only make 1 combination, since {Bob, Alice} and {Alice, Bob} are the same. However in a bitstring of two, I can have 01 and 10, wich are not the same.

Answer 1

"Ordering does not matter in combinations" means that "$0_a 1_a 0_b$" is the same as "$0_b 1_a 0_a$". You have 6 zeros: $0_a,0_b,0_c,0_d,0_e,0_f$ and you have 4 ones: $1_a,1_b,1_c,1_d$ and you have to find how many ways there are to arrange them. If order mattered there would but $10!$ ways.

The order of the the arrangement of the zeros as opposed to the ones does matter, but the order of the specific ones and the specific zeros within a pattern do.

There are $10!$ ways to place the ones and zeros. But as the zeros are interchangible and for any pattern of ones and zero there are $6!$ ways to order the specific zeros within that pattern, so we must divide by the $6!$ ways. Likewise for the ones we must divide by $4!$.

So the answer is ${10 \choose 6}$.

In you anology suppose you had {Alice, Bob} and you also had {Babar, Tantor} and you had to answer "how many ways are there to arrange two people and two elephants".

The answers are:

{Alice,Bob,Babar,Tantor} = {Alice, Bob, Tantor, Babar} = {Bob, Alice, Babar, Tantor} = {Bob, Alice, Tantor, Babar}

{Alice, Babar, Bob, Tantor} = {Alice,Tantor , Bob, Babar}={Bob,Tantor , Alice, Babar}={Bob,Babar , Alice, Tantor}

.... etc....

There are $4! = 24$ ways to arrange the four beings but as we can always switch the two humans and we can always switch the two elephants we get $24/2*2 = 6$ ways:

{(Alice or Bob), (Bob or Alice), (Babar or Tantor), (Tantor or Babar)}

{(Alice or Bob), (Babar or Tantor), (Bob or Alice),(Tantor or Babar)}

{(Alice or Bob), (Babar or Tantor),(Tantor or Babar), (Bob or Alice)}

{(Babar or Tantor), (Alice or Bob), (Bob or Alice), (Tantor or Babar)}

{(Babar or Tantor), (Alice or Bob), (Tantor or Babar), (Bob or Alice)}

{(Babar or Tantor), (Tantor or Babar), (Alice or Bob), (Bob or Alice)}

Answer 2

Choosing a bit like that is the same thing as choosing where are the six $0$.

Which gives you exactly $$\binom {10}6=\frac{10!}{6!(10-6)!}=210$$

possibilities so you are right.

Answer 3

A bit string of length 10 with 6 zeros can be described by a list of the 6 "places" where the zeroes sit. This involves choosing 6 numbers from the 10 possible places, hence 10 choose 6.

Answer 4

Following your first example, choosing a bit string of length $10$ with $6$ zeros, is choosing which six bits among ten are zero.

Answer 5

The answer to this question is $\binom {10} 6$. However, I find this odd. Why would it be a combination in this case and not a permutation?

The "combination" counts the ways to select $6$ places from $10$.   $\tbinom{10}6$

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Alternatively, there are $10!$ permutations of $10$ distinct items, but $6$ and $4$ of these items are not distinct.   There are $6!$ identical ways to arrange the zeros, and $4!$ identical ways to arrange the ones.   So to count using permutations we divide and calculate.   $\frac{10!}{6!~4!}$