MATHHX B

MATHHX B

\(\newcommand{\footnotename}{footnote}\) \(\def \LWRfootnote {1}\) \(\newcommand {\footnote }[2][\LWRfootnote ]{{}^{\mathrm {#1}}}\) \(\newcommand {\footnotemark }[1][\LWRfootnote ]{{}^{\mathrm {#1}}}\) \(\let \LWRorighspace \hspace \) \(\renewcommand {\hspace }{\ifstar \LWRorighspace \LWRorighspace }\) \(\newcommand {\mathnormal }[1]{{#1}}\) \(\newcommand \ensuremath [1]{#1}\) \(\newcommand {\LWRframebox }[2][]{\fbox {#2}} \newcommand {\framebox }[1][]{\LWRframebox } \) \(\newcommand {\setlength }[2]{}\) \(\newcommand {\addtolength }[2]{}\) \(\newcommand {\setcounter }[2]{}\) \(\newcommand {\addtocounter }[2]{}\) \(\newcommand {\arabic }[1]{}\) \(\newcommand {\number }[1]{}\) \(\newcommand {\noalign }[1]{\text {#1}\notag \\}\) \(\newcommand {\cline }[1]{}\) \(\newcommand {\directlua }[1]{\text {(directlua)}}\) \(\newcommand {\luatexdirectlua }[1]{\text {(directlua)}}\) \(\newcommand {\protect }{}\) \(\def \LWRabsorbnumber #1 {}\) \(\def \LWRabsorbquotenumber "#1 {}\) \(\newcommand 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#1.#2.#3\LWRsiunitxEND {\LWRsiunitxprintdecimalsubtwo #1,,\LWRsiunitxENDTWO \ifblank {#2}{}{{\LWRsiunitxdecimal }\LWRsiunitxprintdecimalsubtwo #2,,\LWRsiunitxENDTWO }}\) \(\newcommand {\LWRsiunitxprintdecimal }[1]{\LWRsiunitxprintdecimalsub #1...\LWRsiunitxEND }\) \(\def \LWRsiunitxnumplus #1+#2+#3\LWRsiunitxEND {\ifblank {#2}{\LWRsiunitxprintdecimal {#1}}{\ifblank {#1}{\LWRsiunitxprintdecimal {#2}}{\LWRsiunitxprintdecimal {#1}\unicode {x02B}\LWRsiunitxprintdecimal {#2}}}\LWRsiunitxdistribunit }\) \(\def \LWRsiunitxnumminus #1-#2-#3\LWRsiunitxEND {\ifblank {#2}{\LWRsiunitxnumplus #1+++\LWRsiunitxEND }{\ifblank {#1}{}{\LWRsiunitxprintdecimal {#1}}\unicode {x02212}\LWRsiunitxprintdecimal {#2}\LWRsiunitxdistribunit }}\) \(\def \LWRsiunitxnumpmmacro #1\pm #2\pm #3\LWRsiunitxEND {\ifblank {#2}{\LWRsiunitxnumminus #1---\LWRsiunitxEND }{\LWRsiunitxprintdecimal {#1}\unicode {x0B1}\LWRsiunitxprintdecimal {#2}\LWRsiunitxdistribunit }}\) \(\def \LWRsiunitxnumpm #1+-#2+-#3\LWRsiunitxEND {\ifblank 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{#2}{\LWRsiunitxnume #1eee\LWRsiunitxEND }{\ifblank {#3}{\LWRsiunitxnume #1eee\LWRsiunitxEND \times \LWRsiunitxnume #2eee\LWRsiunitxEND }{\LWRsiunitxnume #1eee\LWRsiunitxEND \times \LWRsiunitxnume #2eee\LWRsiunitxEND \times \LWRsiunitxnume #3eee\LWRsiunitxEND }}}\) \(\newcommand {\num }[2][]{\LWRsiunitxnumx #2xxxxx\LWRsiunitxEND }\) \(\newcommand {\si }[2][]{\mathrm {\gsubstitute {#2}{~}{\,}}}\) \(\def \LWRsiunitxSIopt #1[#2]#3{\def \LWRsiunitxdistribunit {\,\si {#3}}{#2}\num {#1}\def \LWRsiunitxdistribunit {}}\) \(\newcommand {\LWRsiunitxSI }[2]{\def \LWRsiunitxdistribunit {\,\si {#2}}\num {#1}\def \LWRsiunitxdistribunit {}}\) \(\newcommand {\SI }[2][]{\ifnextchar [{\LWRsiunitxSIopt {#2}}{\LWRsiunitxSI {#2}}}\) \(\newcommand {\numlist }[2][]{\text {#2}}\) \(\newcommand {\numrange }[3][]{\num {#2}\ \LWRsiunitxrangephrase \ \num {#3}}\) \(\newcommand {\SIlist }[3][]{\text {#2}\,\si {#3}}\) \(\newcommand {\SIrange }[4][]{\num {#2}\,#4\ \LWRsiunitxrangephrase \ \num {#3}\,#4}\) 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{\morecmidrules }{}\) \(\newcommand {\specialrule }[3]{\hline }\) \(\newcommand {\addlinespace }[1][]{}\) \(\def \LWRsiunitxrangephrase { \protect \mbox {to (numerical range)} }\) \(\def \LWRsiunitxdecimal {.}\)

13.1 Kombinatorik

Kombinatorik handler om, hvor mange forskellige måder noget kan gøres på. Altså hvor mange kombinationer man kan lave. Måske kender i tælletræer fra folkeskolen? Kombinatorik er et spændende emne, men vi skal kun kigge på enkelte begreber, som danner forudsætning for at vi kan forstå den mere avancerede sandsynlighedsregning som kommer senere.

  • Definition 13.1.1
    Vi definerer \(n!\) ved:

    \[n!=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\]

    Tallet \(n!\) kaldes n-fakultet

Ovenstående definition gælder når \(n>1\). Vi definerer \(1!=1\) og \(0!=1\).

  • Eksempel 13.1.1
    Vi udregner \(4!=4\cdot 3\cdot 2\cdot 1=24\).

Øvelse 13.1.1

Bestem ved hovedregning

  • a) \(5!\)

  • b) \(0!\)

Løsning 13.1.1

  • a) \(5!=120\)

  • b) \(0!=1\)

Vi bruger n-fakultet til at bestemme antallet af mulige rækkefølger man kan opstille \(n\) elementer.

  • Eksempel 13.1.2
    Har man 20 elever som gerne vil købe en billet til et fest, så kan man finde ud af, hvor mange muligheder der er for at danne køen til elevcenteret. På den første plads kan der stå 20 elever. For hver at de 20 muligheder er der 19 muligheder for den efterfølgende elev, dvs. \(20\cdot 19\) i alt. For hver af de \(20\cdot 19\) muligheder er der nu 18 muligheder for den næste, dvs. \(20\cdot 19\cdot 18\) i alt osv.

    I sidste ende får vi altså antallet af muligheder til at blive:

    \[20\cdot 19\cdot 18\cdot 17\cdots 4\cdot 3\cdot 2\cdot 1=20!\]

Øvelse 13.1.2

  • a) Bestem antallet af måder man kan opstille 10 i elever i en kø til kantinen.

Løsning 13.1.2

  • a) Der er \(10!=3628800\) måder at lave køen på (overaskende mange hva?)

Vi skal nu se på et lidt mere kompliceret kombinatorisk problem. Forstil jer at der kun er to billetter tilbage til en skolefest og der er 5 elever som gerne vil med. Hvor mange måder er der at fordele billetterne på? En mulighed kunne være:

.
Navn Wan Dennis Jeffrey Daniel Jellow
Får billet \(\times \) \(\times \)

Her får altså Jeffrey og Jellow en billet. Men vi kunne have fordelt de to billetter på mange måde, og vi skal nu se hvordan man finder antallet af måder vi kan fordele billetterne.

  • Definition 13.1.2
    Binomialkoefficienten \(K(n,r)\) er antallet af måder man kan sætte r krydser på \(n\) pladser.

  • Sætning 13.1.1
    Binomialkoefficienten \(K(n,r)\) kan bestemmes ved formlen

    \[K(n,r)=\frac {n!}{r!(n-r)!}\]

  • Eksempel 13.1.3
    Ved hjælp af sætning 13.1.1 kan vi bestemme antallet af måder, vi kan fordele billetterne til skolefesten. Vi har to krydser så \(r=2\) og vi har 5 pladser så \(n=5\). Vi udregner:

    \begin{align*} K(n,r) & =\frac {n!}{r!(n-r)!}\\ & =\frac {5!}{2!(5-2)!}\\ & =\frac {5!}{2!\cdot 3!}\\ & =\frac {5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 3\cdot 2\cdot 1}\\ & =\frac {120}{12}=10 \end{align*} Altså er der 10 måder billetterne kan fordeles på.

Øvelse 13.1.3

  • a) På hvor mange måder, kan man fordele 3 billetter til 6 mennesker?

Løsning 13.1.3

  • a) Der er 20 muligheder.

Øvelse 13.1.4 (Svær)

En elev fra Niels Brock skal holde fødselsdagsfest for klassen.

Eleven har 8 store kasser med slik. I hver kasse er der netop en type slik og der er ikke to kasser med samme type.

Eleven vil gerne lave slikposer med 4 forskellige stykker slik i hver, og der må ikke være to ens poser. Der skal være en pose til hver elev.

  • a) Kan det lade sig gøre?

Løsning 13.1.4

  • a) JA!