There are 20 bags in a room, each one with a 5 dollar and 7 dollar bill. Each bag contains a bill of a different color (the bills from different bags are distinguishable). You can select either 0 or 1 bill from each bag. How many ways are there to pay 17 dollars by selecting bills from these bags. Paying with bills from different bags is considered different.
", "hints_md": "Use [Generating Functions (GF)](https://artofproblemsolving.com/wiki/index.php/Generating_function). You'll also need some knowledge on the [Maclaurin Series](https://artofproblemsolving.com/wiki/index.php/Taylor_polynomial#Maclaurin_polynomial) and the [Binomial Theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem).\n\nMaclaurin & Binomial Theorem TLDR for this problem ($n \\in \\mathbb{C}$): $$(1+x)^n = 1+nx+\\frac{n(n-1)}{2!}x^2+\\frac{n(n-1)(n-2)}{3!}x^3+\\cdots=\\sum\\limits_{k=0}^\\infty \\frac{n(n-1)\\cdots(n-k+1)x^k}{k!}. $$", "hints_html": "Use Generating Functions (GF). You’ll also need some knowledge on the Maclaurin Series and the Binomial Theorem.
Maclaurin & Binomial Theorem TLDR for this problem (n \\in \\mathbb{C}): (1+x)^n = 1+nx+\\frac{n(n-1)}{2!}x^2+\\frac{n(n-1)(n-2)}{3!}x^3+\\cdots=\\sum\\limits_{k=0}^\\infty \\frac{n(n-1)\\cdots(n-k+1)x^k}{k!}.
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