woowheedr

Answered

2022-07-08

Find the number of solutions in integers to ${n}_{1}+{n}_{2}+{n}_{3}+{n}_{4}=12$ satisfying $0\le {n}_{1}\le 4,0\le {n}_{2}\le 5,0\le {n}_{3}\le 8,\text{and}0\le {n}_{4}\le 9$.

Answer & Explanation

grubijanebb

Expert

2022-07-09Added 10 answers

Step 1

Let A denote nonnegative solutions to ${n}_{1}+{n}_{2}+{n}_{3}+{n}_{4}=12$ and ${A}_{i}$ denote those solutions where the upper bound on ${n}_{i}$ is violated, then inclusion/exclusion gives $|\overline{{A}_{1}\cup {A}_{2}\cup {A}_{3}\cup {A}_{4}}|=|A|-(|{A}_{1}|+|{A}_{2}|+|{A}_{3}|+|{A}_{4}|)+(|{A}_{1}\cap {A}_{2}|+\cdots )-\cdots \phantom{\rule{0ex}{0ex}}={\textstyle (}\genfrac{}{}{0ex}{}{12+4-1}{12}{\textstyle )}-({\textstyle (}\genfrac{}{}{0ex}{}{7+4-1}{7}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{6+4-1}{6}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{3+4-1}{3}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{2+4-1}{2}{\textstyle )})+{\textstyle (}\genfrac{}{}{0ex}{}{1+4-1}{1}{\textstyle )}$ where e.g. $|{A}_{1}|={\textstyle (}\genfrac{}{}{0ex}{}{7+4-1}{7}{\textstyle )}$

Step 2

Because the solutions there can be recast as solutions to $({m}_{1}+5)+{n}_{2}+{n}_{3}+{n}_{4}=12$ with ${m}_{1},{n}_{i}\ge 0$ or ${m}_{1}+{n}_{2}+{n}_{3}+{n}_{4}=7$, and omitted terms are because they are empty (i.e. they reduce to a sum of nonnegative integers to a negative integer).

Let A denote nonnegative solutions to ${n}_{1}+{n}_{2}+{n}_{3}+{n}_{4}=12$ and ${A}_{i}$ denote those solutions where the upper bound on ${n}_{i}$ is violated, then inclusion/exclusion gives $|\overline{{A}_{1}\cup {A}_{2}\cup {A}_{3}\cup {A}_{4}}|=|A|-(|{A}_{1}|+|{A}_{2}|+|{A}_{3}|+|{A}_{4}|)+(|{A}_{1}\cap {A}_{2}|+\cdots )-\cdots \phantom{\rule{0ex}{0ex}}={\textstyle (}\genfrac{}{}{0ex}{}{12+4-1}{12}{\textstyle )}-({\textstyle (}\genfrac{}{}{0ex}{}{7+4-1}{7}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{6+4-1}{6}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{3+4-1}{3}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{2+4-1}{2}{\textstyle )})+{\textstyle (}\genfrac{}{}{0ex}{}{1+4-1}{1}{\textstyle )}$ where e.g. $|{A}_{1}|={\textstyle (}\genfrac{}{}{0ex}{}{7+4-1}{7}{\textstyle )}$

Step 2

Because the solutions there can be recast as solutions to $({m}_{1}+5)+{n}_{2}+{n}_{3}+{n}_{4}=12$ with ${m}_{1},{n}_{i}\ge 0$ or ${m}_{1}+{n}_{2}+{n}_{3}+{n}_{4}=7$, and omitted terms are because they are empty (i.e. they reduce to a sum of nonnegative integers to a negative integer).

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