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"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?

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  • $\begingroup$ $10!$ is correct. If the strings are distinct, the compositions are distinct as well. $\endgroup$ – Peter May 16 '18 at 12:18
  • $\begingroup$ That's right, but how do I choose those bits on those 10 segments $\endgroup$ – Michał May 16 '18 at 12:20
  • $\begingroup$ I assumed that the $10$ distinct strings are already given. $\endgroup$ – Peter May 16 '18 at 12:26
  • $\begingroup$ 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? $\endgroup$ – Michał May 16 '18 at 12:28
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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?

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  • $\begingroup$ 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? $\endgroup$ – Michał May 16 '18 at 12:23
  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig May 16 '18 at 12:30
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    $\begingroup$ So my anwser is good but without $10!$ right? I get it now $\endgroup$ – Michał May 16 '18 at 12:31
  • $\begingroup$ Yes, that is correct. $\endgroup$ – N. F. Taussig May 16 '18 at 12:32

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