Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You probably mean $6^5$ different results? – gt6989b Feb 11 '14 at 18:28
@gt6989b Yes, my mistake. – Wyvern666 Feb 11 '14 at 18:28
How do you get that expression that evaluates to $651$? – ShreevatsaR Feb 11 '14 at 18:32
@ShreevatsaR Trying to simplifiy the original equation. I assign one element to the five variables, an then decrease the right side: $20-5=15$ – Wyvern666 Feb 11 '14 at 18:35
@ShreevatsaR…*+%5Cbinom%7B13%7D%7B9%7D%2B%5Cbinom%7B5%7D%7B2%7D+*%5Cbinom%7B7%7D%7B3%7D shows this evaluates to 651 – gt6989b Feb 11 '14 at 18:46
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thanks for your answer, im going to accept it. I have recently find the solution in the book, and i was right. – Wyvern666 Feb 11 '14 at 19:03

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 $f(x)^5$: $$ \begin{split} \left[x^{15}\right]\frac{\left(1-x^6\right)^5}{(1-x)^5} &= \left[x^{15}\right]\frac{1-5x^6+10x^{12}}{(1-x)^5} \\ &= \left[ \left[x^{15}\right] -5\left[x^{9}\right] + 10 \left[x^{3}\right]\right] \frac{1}{(1-x)^5}\\ &= 3876 - 5 \cdot 715 + 10 \cdot 35 \\ &= 651. \end{split} $$

share|cite|improve this answer
I dont know what are generating functions. And what means 336? – Wyvern666 Feb 11 '14 at 18:46
$336$ would be the numerator for your probability fraction. – John Habert Feb 11 '14 at 18:47
@JohnHabert Well, is very far from the supposed solution. – Wyvern666 Feb 11 '14 at 18:58
@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. – John Habert Feb 11 '14 at 19:03
@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 :) – gt6989b Feb 11 '14 at 19:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.