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This is supposed to be a Inclusion-Exclusion problem.

We have $6^5=7776$ different results.

Now, with the Inclusion-Exclusion principle i resolve the number of solutions for the equation:

$d_1+d_2+d_3+d_4+d_5=20 , 1 \leq d_i \leq 6 ,\forall 1 \leq i \leq 5$

That, i think, is equivalent to resolve:

$d_1+d_2+d_3+d_4+d_5=15 , 0 \leq d_i \leq 5 ,\forall 1 \leq i \leq 5$

This is:

$\binom{19}{15}-\binom{5}{1} \binom{13}{9}+\binom{5}{2} \binom{7}{3}=651$

So mi answer is: $\frac{651}{7776}$.

However, the "correct answer" is: $\frac{116}{7776}$. Whats the problem with mi reasoning?

EDIT I found the solutions for my problem in the book, the result is: $\frac{651}{7776}$.

I was right, but the solution from mi teacher was wrong. Thats why the confusion.

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  • $\begingroup$ You probably mean $6^5$ different results? $\endgroup$
    – gt6989b
    Feb 11, 2014 at 18:28
  • $\begingroup$ @gt6989b Yes, my mistake. $\endgroup$
    – Wyvern666
    Feb 11, 2014 at 18:28
  • $\begingroup$ How do you get that expression that evaluates to $651$? $\endgroup$ Feb 11, 2014 at 18:32
  • $\begingroup$ @ShreevatsaR Trying to simplifiy the original equation. I assign one element to the five variables, an then decrease the right side: $20-5=15$ $\endgroup$
    – Wyvern666
    Feb 11, 2014 at 18:35
  • $\begingroup$ @ShreevatsaR wolframalpha.com/input/…*+%5Cbinom%7B13%7D%7B9%7D%2B%5Cbinom%7B5%7D%7B2%7D+*%5Cbinom%7B7%7D%7B3%7D shows this evaluates to 651 $\endgroup$
    – gt6989b
    Feb 11, 2014 at 18:46

2 Answers 2

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Using generating functions, we get a generating function of $$f(x) = 1+x+\ldots+x^5 = \frac{1-x^6}{1-x}$$ for each variable so we are now looking for the coefficient of $x^{15}$ in $f(x)^5$: $$ \begin{split} \left[x^{15}\right]\frac{\left(1-x^6\right)^5}{(1-x)^5} &= \left[x^{15}\right]\frac{1-5x^6+10x^{12}}{(1-x)^5} \\ &= \left[ \left[x^{15}\right] -5\left[x^{9}\right] + 10 \left[x^{3}\right]\right] \frac{1}{(1-x)^5}\\ &= 3876 - 5 \cdot 715 + 10 \cdot 35 \\ &= 651. \end{split} $$

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  • $\begingroup$ I dont know what are generating functions. And what means 336? $\endgroup$
    – Wyvern666
    Feb 11, 2014 at 18:46
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    $\begingroup$ $336$ would be the numerator for your probability fraction. $\endgroup$ Feb 11, 2014 at 18:47
  • $\begingroup$ @JohnHabert Well, is very far from the supposed solution. $\endgroup$
    – Wyvern666
    Feb 11, 2014 at 18:58
  • $\begingroup$ @Wyvern666 I'm not making any claim as to the accuracy of the method. I'm just answering your question about what the $336$ means. If I could answer accurately myself, I would. I leave it to those with more knowledge. $\endgroup$ Feb 11, 2014 at 19:03
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    $\begingroup$ @vonbrand actually checked the expansion, your coefficients come out the same, $\binom{19}{4} = 3876, \binom{13}{4} = 715, \binom{7}{4} = 35$. Wolfram Alpha doesn't lie :) $\endgroup$
    – gt6989b
    Feb 11, 2014 at 19:56
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You can calculate this by finding the coefficient on $x^{20}$ in the polynomial $$ \left(\frac{1}{6}x+\frac{1}{6}x^2+\frac{1}{6}x^3+\frac{1}{6}x^4+\frac{1}{6}x^5+\frac{1}{6}x^6 \right)^5 $$ Using PARI/GP, I get that this polynomial is $$\frac{1}{7776} x^{30} + \frac{5}{7776} x^{29} + \frac{5}{2592} x^{28} + \frac{35}{7776} x^{27} + \frac{35}{3888} x^{26} + \frac{7}{432} x^{25} + \frac{205}{7776} x^{24} + \frac{305}{7776} x^{23} + \frac{35}{648} x^{22} + \frac{5}{72} x^{21} + \frac{217}{2592} x^{20} + \frac{245}{2592} x^{19} + \frac{65}{648} x^{18} + \frac{65}{648} x^{17} + \frac{245}{2592} x^{16} + \frac{217}{2592} x^{15} + \frac{5}{72} x^{14} + \frac{35}{648} x^{13} + \frac{305}{7776} x^{12} + \frac{205}{7776} x^{11} + \frac{7}{432} x^{10} + \frac{35}{3888} x^9 + \frac{35}{7776} x^8 + \frac{5}{2592} x^7 + \frac{5}{7776} x^6 + \frac{1}{7776} x^5 $$ This shows that your value, $\frac{651}{7776}=\frac{217}{2592}$ is correct.

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  • $\begingroup$ Thanks for your answer, im going to accept it. I have recently find the solution in the book, and i was right. $\endgroup$
    – Wyvern666
    Feb 11, 2014 at 19:03

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