سوال ۱۱۳ کنکور ارشد علوم کامپیوتر ۱۴۰۵

Reduce the system. Since $x_1+x_2+x_3=5$, substitute this into the second equation:

$5 + x_4 + x_5 = 20 \;\Rightarrow\; x_4 + x_5 = 15.$

The system is therefore equivalent to two independent stars-and-bars problems on disjoint variable sets:

• $x_1+x_2+x_3=5$ over non-negative integers: $\binom{5+3-1}{3-1}=\binom{7}{2}=21$ solutions.
• $x_4+x_5=15$ over non-negative integers: $\binom{15+2-1}{2-1}=\binom{16}{1}=16$ solutions.

Multiplication principle. The two blocks share no variables, so the total number of solutions is

$\binom{7}{2}\cdot\binom{16}{1}=21\cdot 16 = 336,$

which is option 4.

Pitfall: The two equations share $x_1,x_2,x_3$, so they are not independent. Treating the second one as a fresh equation $x_1+\dots+x_5=20$ gives $\binom{24}{4}$ and produces trap options 2 and 3; adding the two counts $21+16=37$ gives trap option 1. Eliminate the shared part first, then multiply.