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I need some help with combinatorics. I have to count all the 10 bit words that contain at least 3 '1' and 3 '0', so I guess that the words would be something like this:

$$111 000 xxx x$$

The problem is that I don't really know how to determine which formula I need to use. Are these permutations with repetition, or are they variations? Becase the order is important I think that they most certainly are not combinations.

Also, I know that there are $2^n$ different 10-bit words, so there can not be more than $2^n$ such words.

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  • $\begingroup$ Are you aware of the multinomial coeff.? $\endgroup$ – Studentmath Oct 7 '14 at 11:00
  • $\begingroup$ I only know how to use the binomial theorem $\endgroup$ – Mark Oct 7 '14 at 11:11
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Do it by complimentary condition. How many $10$ bit words contain at most $2$ $1$s or at most $2$ $0$s?

Consider case of $0$s. The other calculation is similar.

Case 1: $0$ $0$s. Exactly $1$ such word: ${10\choose 0} = 1$

Case 2: $1$ $0$s. Choose a position for $0$: ${10 \choose 1} = 10$.

Case 3: $2$ $0$s. Choose $2$ positions for $0$s: ${10\choose 2} = 45$.

Consider same events for $1$, add those, and subtract from $2^{10}$ to get required number.

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  • $\begingroup$ In fact, it is the exactly same thing, so can we just add these results $(1 + 10 + 45)$ and multiply it by $2$ and in the end get $112$, right? $\endgroup$ – Mark Oct 7 '14 at 11:16
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    $\begingroup$ Yes. Don't forget to subtract that from $2^{10}$ though. $\endgroup$ – taninamdar Oct 7 '14 at 12:01

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