# Stars and Bars problem involving odd restriction, and equal or greater than restriction.

I just had this question in an exam and was unsure how to complete some parts using the Stars and Bars method.

Problem as follows:

How many solutions has the equation:

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 78$

If $x_1, x_2, x_3, x_4, x_5, x_6$ are non-negative integers and:

a) there are no further restrictions

b) all of $x_1, x_2, x_3, x_4, x_5, x_6$ are odd numbers

c) at least one of $x_1, x_2, x_3, x_4, x_5, x_6$ is greater than or equal to 30.

My attempts if they are worth anything, I believe I'm not familar enough with the method which lead to me misusing it:

a) This one was straight foward, I got $\binom{83}{5}$, by defintion of the method.

b) I'm certain I got this part wrong. I used my result from part (a), and minused it by the number of solutions where $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 78$ is even. I.e. where we $x_1, x_2, x_3, x_4, x_5, x_6$ represent two stars such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 39$.

So the final calculation was $\binom{83}{5}$ - $\binom{44}{5}$, but this doesn't make numerical sense. $\ddot\frown$

c) Not too sure about this one either. I tried 'reserving' 30 stars such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 48$, so $\binom{53}{5}$.

Similarly for the other case where there is some $x_i$ and $x_j$ are both greater than 30 so I 'reserve' 60 stars such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 18$ so $\binom{18}{5}$.

Then I added the two together such that $\binom{18}{5} + \binom{53}{5}$

• In c) the reserved $30$ can be added to $6$ candidates, giving a factor $6$ for the number of possibilities. Secondly the twice reserved $30$ gives likewise a factor $\binom{6}{2}=15$. Thirdly (in my view the most tricky) there is a double counting. A possibility where $30$ is added can be the same as one where $30$ is added twice. – drhab Nov 18 '14 at 10:56

Find the number of solutions of $z_{1}+\cdots+z_{6}=36$ under no restrictions.
This comes to the same as finding the number of solutions of $y_{1}+\cdots+y_{6}=72$ under condition that all numbers are even.
And on its turn it comes to the same as finding the number of solutions of $x_{1}+\cdots+x_{6}=78$ under condition that all numbers are odd.
This by taking $y_i=2z_i$ and $x_{i}=y_{i}+1=2z_{i}+1$.