How many different strings of length $100$ may be composed of $10$ different $10$ position binary numbers?

"How many different strings of length $100$ may be composed of $10$ different $10$ position binary numbers?"

So this series would be divided into $10$ segments of $10$ bits. Maximum number of options at one segment is $2^{10}$, my idea is that I choose $2^{10}$ for each of those segments and therefore do $10!$ to permute them, although in this case those strings wouldn't differ from each other, any idea how to solve this problem?

• $10!$ is correct. If the strings are distinct, the compositions are distinct as well. – Peter May 16 '18 at 12:18
• That's right, but how do I choose those bits on those 10 segments – Michał May 16 '18 at 12:20
• I assumed that the $10$ distinct strings are already given. – Peter May 16 '18 at 12:26
• No I need to choose those 10 strings of 10 bits, as I anwsered under below anwser : " So ($2^{10} * (2^{10} -1 ) * (2^{10} -2 ) * (2^{10} -3 ) * (2^{10} -4 ) *(2^{10} -5 ) *(2^{10} -6) *(2^{10} -7 ) *(2^{10} -8 ) *(2^{10} -9 )* 10!$ Is that right? – Michał May 16 '18 at 12:28

There are $2^{10}$ ways to fill the first $10$ digits. Since the next ten digits must differ from the first ten digits, they can be selected in $2^{10} - 1$ ways. Can you continue?
• So ($2^{10} * (2^{10} -1 ) * (2^{10} -2 ) * (2^{10} -3 ) * (2^{10} -4 ) *(2^{10} -5 ) *(2^{10} -6) *(2^{10} -7 ) *(2^{10} -8 ) *(2^{10} -9 )* 10!$ Is that right? – Michał May 16 '18 at 12:23
• Not quite. You have already accounted for the order by choosing which string is first, which is second, and so forth, so there is no need to multiply by $10!$. To see this, consider the related problem of finding the number of permutations of three different digits selected from the ten decimal digits. We can select the first digit in $10$ ways. For each such choice, we can select the second digit in $9$ ways. For each such choice, we can select the third digit in eight ways. Therefore, there are $10 \cdot 9 \cdot 8$ such strings. – N. F. Taussig May 16 '18 at 12:30
• So my anwser is good but without $10!$ right? I get it now – Michał May 16 '18 at 12:31