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What I have so far:

Goal: Using the inclusion exclusion I want to find $|\overline A_{1}\cap \overline A_{2} \cap \overline A_{3} \cap \overline A_{4}| = |U| - S_{1} + S_{2} - S_{3} + S_{4}$

$S_{k} = \sum |\overline A_{i1}\cap \overline A_{i1} \cap ... \overline A_{ik}|$

I have incremented the values of i by 5 so that the range can start from zero like this:

$x_{1}+x_{2}+ x_{3} + x_{4} = 24$ with $0\leq x_{i} \leq 15$

For $|U|$ I have used to "stars and bars technique":

$|U| = \binom{r+n-1}{r} = \binom{24+4-1}{3} $

The Answer

... I am studying for a test (this is a practice question) and my professor has provided a solution that says: $\binom{42}{39} - \binom{4}{1} \binom{26}{23}+\binom{4}{2}\binom{10}{7}$

So I don't think I am on the right track if the universal set $|U| = \binom{42}{39}$. Any tips would be great thanks in advance.

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    $\begingroup$ Well, one thing if you are incrementing the four $x_i$ by $5$ then total needs to increment by $4*5$. You only incremented it by $5$. $\endgroup$ – fleablood Nov 19 '18 at 0:38
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I have incremented the values of i by 5 so that the range can start from zero like this: $ x_1+x_2+x_3+x_4=24$ with $0≤x_i≤15$

Are you sure this is correct? You might need to check that inequality and the equation preceding it.

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    $\begingroup$ After fixing the inequality to $0≤x_i+5≤15$, you should find that $x_1+x_2+x_3+x_4 = 39$. Then using your formula for $|U|$, you should find the answer. $\endgroup$ – LeNoir Nov 19 '18 at 0:55
  • $\begingroup$ Thanks leNoir that fixed it! $\endgroup$ – inventiveOak4230 Nov 19 '18 at 4:47

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