How many solutions are there to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 29$$ where $x_i , i = 1, 2, 3,4,5, 6$ are nonnegative integers such that
a) $x_i > 1$ for $i = 1, 2, 3, 4, 5, 6$?
c) $x_1 \le 5$?
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2$\begingroup$ Have you tried writing all of them down? Also, is there a condition (b)? $\endgroup$– Prahlad VaidyanathanDec 1, 2013 at 5:55
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1$\begingroup$ Use the stars and bars. $\endgroup$– nomenDec 1, 2013 at 5:55
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2 Answers
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Throw $2$ balls in each box, to make sure i) is satisfied, and then solve $x_1+x_2+..+x_6=17$ for each of $x_1=1,2,3$ separately, using the formula $(n+k-1)C(k-1)$, where $n$ is the total sum, and $k$ is the number of terms in the sum.
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For (a) replace $x_k$ by $y_k=x_k-2$; then you’re looking for solutions in non-negative integers to
$$y_1+y_2+y_3+y_4+y_5+y_6=17\;,$$
a standard stars and bars problem. For (c) count the solutions in non-negative integers to
$$x_1+x_2+x_3+x_4+x_5+x_6=29\;,$$
and subtract those that have $x_1\ge 6$; you can count those in much the same way that I suggested solving (a).
answered Dec 1, 2013 at 6:02