Weitere Übungen zur ersten Lektion
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Ein Restaurant hat eine Speisekarte mit $\\var{a}$ Vorspeisen, $\\var{b}$ Hauptgerichten und $\\var{c}$ Desserts.
\nWie viele Möglichkeiten gibt es, sich als Gast ein Menü mit genau einer Vorspeise, einem Hauptgericht und einem Dessert zusammenzustellen?
\n", "advice": "
Die Anzahl der Möglichkeiten ergibt sich gemäß Produktregel als $\\var{a}\\cdot \\var{b}\\cdot \\var{c}=\\var{a*b*c}$
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "In einem Dorf verkaufen zwei Geschäfte Sweatshirts. Das erste Geschäft hat $\\var{a}$ Modelle in jeweils $\\var{c}$ Farben. Das zweite Geschäft hat $\\var{b}$ Modelle in $\\var{d}$ Farben. Die Modelle im ersten und zweiten Geschäft sind alle voneinander verschieden. Wie viele verschiedene Sweatshirts kann man in diesem Dorf kaufen?
", "advice": "Es gibt $\\var{a} \\cdot \\var{c}=\\var{a*c}$ verschiedene Sweatshirts im ersten Geschäft und $\\var{b} \\cdot \\var{d}=\\var{b*d}$ im zweiten Geschäft. Beide Anzahlen dürfen addiert werden (disjunkte Mengen), also gibt es $\\var{ans}$ verschiedene Sweatshirts zu kaufen.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Ein Autohandel beschäftigt $\\var{p}$ Verkäufer*innen. Ein*e Verkäufer*in erhält €$\\,100$ Bonus für jedes verkaufte Auto.
\nGestern wurden $\\var{b}$ Autos verkauft.
\nAuf wie viele verschiedene Arten kann der auszuzahlende Bonus von €$\\,\\var{b*100}$ unter den $\\var{p}$ Verkäufer*innen aufzuteilen sein?
", "advice": "Wir betrachten die Reihenfolge der Autoverkäufe für unwichtig, allerdings die Anzahl der je Verkäufer*in verkauften Autos für wichtig.
\nDies ist ein Anwendungsfall für die Kombinationen mit Wiederholung, bzw. im Fächer-Teilchen-Modell: Verteilung von $k=\\var{b}$ Teilchen (Autos) auf $n=\\var{p}$ Fächer (Verkäufer*innen), wobei Mehrfachbelegungen erlaubt sind und die Teilchen ununterscheidbar (jedes Auto gibt denselben Bonus von €$\\,100$). Die Anzahl verschiedener Aufteilungen ist demnach:
\n\\[\\binom{\\var{p}+\\var{b}-1}{\\var{b}}=\\var{ans}\\]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Es sei $K$ eine Menge von $\\var{sum}$ Kindern, $M$ die Teilmenge der Mädchen mit $|M|=\\var{m}$, $B$ die Teilmenge der Burschen mit $|B|=\\var{b}$.
", "advice": "(a) Für diese Frage ist es egal, ob es Burschen oder Mädchen sind. Es sind also aus $n=\\var{sum}$ Elementen $k=\\var{(sum+1)/2}$ ohne Wiederholung und ohne Beachtung der Reihenfolge, d.h. Kombinationen (ohne Wiederholung), also $\\binom{\\var{sum}}{\\var{(sum+1)/2}}=\\var{a_result}$.
\n\n\n
(b) Für diese Frage sind die Teilgruppen \"Mädchen\" und \"Burschen\" getrennt zu betrachten, sonst gilt daselbe wie bei (a): Innerhalb der Gruppen erfolgt eine Auswahl ohne Wiederholung und ohne Beachtung der Reihenfolge, man erhält also $\\binom{\\var{m}}{\\var{mini}}$ für die Mädchen und $\\binom{\\var{b}}{\\var{mini}}$ für die Burschen, diese beiden Zahlen sind zu multiplizieren (Entscheidungen sind keine Alternativen und hängen nicht voneinander ab), was $\\var{b_result}$ ergibt.
\n(c) Es gibt 1 Möglichkeit, alle Mädchen auszuwählen, diese ist dann noch mit den insgesamt (alle Teilmengen) $2^{\\var{b}}=\\var{c_result}$ Möglichkeiten zu multiplizieren, $0,1,2,\\ldots \\var{b}$ Burschen auszuwählen.
\n(d) Genau die Hälfte aller Teilmengen enthält weniger als $\\var{(sum+1)/2}$, genau die Hälfte mindestens $\\var{(sum+1)/2}$ Kinder. Die Anzahl aller Teilmengen einer $\\var{sum}$-elementigen Menge ist $2^{\\var{sum}}$, die Hälfte davon ist $2^{\\var{sum-1}}=\\var{d_result}$.
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Wie viele Teilmengen von $K$ enthalten genau $\\var{mini}$ Mädchen und genau $\\var{mini}$ Burschen?
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