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Pe: 2020-04-28T20:30:38+03:00 In: In:

fisa-combinatorica pls help

fisa-combinatorica

pls help

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  1. Menim

    Menim

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    1.Trebuie sa gasim numarul de moduri in care se pot aranja 5 elemente. Aceasta este chiar definitia permutarilor, deci avem P_5=5!=120 de moduri.

    2.Multimea A:
    Orice numar reprezinta o permutare a multimii A. Avem deci P_4=4!=24 numere.
    Multimea B:
    Prima cifra a numarului nu poate fi 0. Ne raman deci 3 variante pentru aceasta. Dupa ce am ales prima cifra, raman 3 posibilitati pentru celelalte 3 cifre. Ca la multimea A, alegerea acestor ultime 3 cifre reprezinta o permutare, deci avem P_3=3!=6 variante. Considerand si cele 3 variante pentru prima cifra, avem 3*6=18 variante.

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  2. Menim

    Menim

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    3.a)3!+4!=6+24=30
    5!+3!=120+6=126
    b)6!-5!=6*5!-5!=5*5!=5*120=600
    10!-9!=10*9!-9!=9*9!
    c)\frac{124!}{125!}=\frac{124!}{124!\cdot 125}=\frac1{125}
    \frac{2011!}{2010!}=\frac{2010!\cdot 2011}{2010!}=2011
    d)\frac{n!}{(n+2)!}=\frac{n!}{n!\cdot(n+1)(n+2)}=\frac1{(n+1)(n+2)}
    \frac{n!}{(n-2)!}=\frac{(n-2)!(n-1)n}{(n-2)!}=n(n-1)
    e)\frac{(2n+1)!}{(2n-2)!}=\frac{(2n-2)!(2n-1)2n(2n+1)}{(2n-2)!}=2n(2n-1)(2n+1)
    \frac{(2k-4)!}{(2k-2)!}=\frac{(2k-4)!}{(2k-4)!(2k-3)(2k-2)}=\frac1{(2k-3)(2k-2)}
    f)\frac{(2n+m)!}{(2n+m+1)!}=\frac{(2n+m)!}{(2n+m)!\cdot(2n+m+1)}=\frac1{2n+m+1}
    g)\frac{(2-n)!}{(3-n)!}=\frac{(2-n)!}{(2-n)!(3-n)}=\frac1{3-n}
    \frac{(5-n)!}{(7-n)!}=\frac{(5-n)!}{(5-n)!(6-n)(7-n)}=\frac1{(6-n)(7-n)}

    4.a)\frac1{2!}-\frac2{3!}=\frac12-\frac26=\frac36-\frac26=\frac16
    \frac1{5!}-\frac2{4!}=\frac1{5!}-\frac{2\cdot5}{5!}=\frac1{5!}-\frac{10}{5!}=\frac{-9}{5!}=\frac{-9}{120}
    b)\frac1{0!}-\frac2{1!}=\frac11-\frac21=1-2=-1

    c)\frac{1!+2!+3!}{9!(10!-8!)}:\frac2{7!}=\frac{1+2+6}{7!\cdot8\cdot9(10!-8!)}\cdot\frac{7!}2=\frac{9}{8\cdot 9(10!-8!)}\cdot\frac12=\frac1{8(8!\cdot9\cdot10-8!)}\cdot\frac12=\frac1{16\cdot(8!\cdot90-8!)}=\frac1{16\cdot89\cdot8!}
    \frac{1!+2!+3!+4!}{5!(9!-8!)}:\frac{10}{8\cdot8!}=\frac{1+2+6+24}{5!(9\cdot8!-8!)}\cdot\frac{8\cdot8!}{10}=\frac{33}{5!\cdot8\cdot8!}\cdot\frac{8\cdot8!}{10}=\frac{33}{5!\cdot10}

    d)1\cdot1!+2\cdot2!+3\cdot3!-4\cdot4!=1\cdot1+2\cdot2+3\cdot6-4\cdot24=1+4+18-96=23-96=-73
    1\cdot1!+2\cdot2!+3\cdot3!+4\cdot4!=1\cdot1+2\cdot2+3\cdot6+4\cdot24=1+4+18+96=23+96=119

    5.Simplificam mai intai fiecare termen al multimii.
    0!=1
    2!-1!=2-1=1
    \frac{5\cdot4!\cdot4}{2!\cdot5!}=\frac{5!\cdot4}{2\cdot5!}=\frac42=2
    Observam ca primele 2 elemente sunt de fapt egale. Multimea A devine:
    A=\left \{ 1, 2 \right \}
    Aceasta multime are 2 elemente, deci 2!=2 permutari, care sunt \left \{ 1, 2 \right \} si \left \{ 2, 1 \right \}.

    6.a)Functia factorial este o functie strict crescatoare, deci ecuatia n!=24 are cel mult o solutie. Observam ca 4!=24, deci solutia ecuatiei este n=4.
    b)\frac{(n+2)!}{(n+1)!}=7
    \frac{(n+2)(n+1)!}{(n+1)!}=7
    n+2=7
    n=5
    c)\frac{(2n+3)!}{(2n+1)!}=20
    \frac{(2n+1)!(2n+2)(2n+3)}{(2n+1)!}=20
    (2n+2)(2n+3)=20
    2(n+1)(2n+3)=20
    (n+1)(2n+3)=10
    2n^2+3n+2n+3=10
    2n^2+5n-7=0
    Aceasta ecuatie are solutiile 1 si -7/2. A 2-a solutie nu convine deoarece factorialul este definit doar pe numere naturale.
    d)\frac{(n+1)!}{(n-3)!}=\frac{30(n+1)!}{(n-1)!}
    (n+1)! nu poate fi 0, deci putem imparti prin (n+1)!:
    \frac1{(n-3)!}=\frac{30}{(n-1)!}
    \frac1{(n-3)!}=\frac{30}{(n-3)!(n-2)(n-1)}
    Inmultim cu (n-3)!:
    1=\frac{30}{(n-2)(n-1)}
    (n-2)(n-1)=30
    Rezolvand aceasta ecuatie de gradul 2, obtinem solutia n=7.
    e)\frac{P_{n+3}}{P_{n+1}}=56
    \frac{(n+3)!}{(n+1)!}=56
    (n+2)(n+1)=56
    Rezolvand ecuatia de gradul 2, obtinem solutia n=6.

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  3. Menim

    Menim

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    7.a)\frac{P_{x+7}}{P_{x+8}}\geq\frac1{16}
    \frac{(x+7)!}{(x+8)!}\geq\frac1{16}
    \frac{(x+7)!}{(x+7)!\cdot(x+8)}\geq\frac1{16}
    \frac1{x+8}\geq\frac1{16}
    x+8\leq16
    x\leq8
    b)\frac{(n+1)!}{(n-1)!}\leq72
    \frac{(n-1)!\cdot n\cdot (n+1)}{(n-1)!}\leq72
    n(n+1)\leq72
    n este numar natural nenul, deci strict pozitiv. Observam ca pentru n>0, partea stanga a inegalitatii este strict crescatoare. Observam deasemenea ca pt n=8, partea stanga iar fix valoarea 72. Rezulta ca solutiile cautate sunt 1, 2, 3, 4, 5, 6, 7, 8.

    8.\begin{cases} P_{x-y}=3P_{x-y-1} \\ P_{2x-y+1}=8P_{2x-y} \end{cases}
    \begin{cases} (x-y)!=3(x-y-1)! \\ (2x-y+1)!=8(2x-y)! \end{cases}
    \begin{cases} (x-y-1)!(x-y)=3(x-y-1)! \\ (2x-y)!(2x-y+1)=8(2x-y)! \end{cases}
    \begin{cases} x-y=3 \\ 2x-y+1=8 \end{cases}
    Scadem prima ecuatie din a 2-a:
    2x-y+1-(x-y)=8-3
    2x-y+1-x+y=5
    x+1=5
    x=4
    Inlocuim x=4 in prima ecuatie din sistem:
    4-y=3
    y=1

    9.nP_n=n\cdot n!=(n+1-1)\cdot n!=(n+1)n!-n!=(n+1)!-n!=P_{n+1}-P_{n}

    a)\sum_{k=1}^{n} k\cdot k!=\sum_{k=1}^{n} k\cdot P_k=\sum_{k=1}^{n}(P_{k+1}-P_k)=-\sum_{k=1}^{n}(P_k-P_{k+1})=-((P_1\cancel{-P_2})+(\cancel{P_2}\cancel{-P_3})+(\cancel{P_3}\cancel{-P_4})+...+(\cancel{P_n}-P_{n+1}))=-(P_1-P_{n+1})=P_{n+1}-P_1=(n+1)!-1!=(n+1)!-1

    b)\sum_{k=1}^{n} k!(k+1)^2=\sum_{k=1}^{n} k!(k+1)\cdot(k+1)=\sum_{k=1}^{n} (k+1)!(k+1)=\sum_{k=1}^{n} (k+1)P_{k+1}=\sum_{k=1}^{n}(P_{k+2}-P_{k+1})=-\sum_{k=1}^{n}(P_{k+1}-P_{k+2})=-(P_2-P_3+P_3-P_4+...+P_{n+1}-P_{n+2})=-(P_2-P_{n+2})=P_{n+2}-P_2=(n+2)!-2!=(n+2)!-2

    10.\frac{n}{P_{n+1}}=\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac1{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}=\frac1{P_n}-\frac1{P_{n+1}}

    a)\sum_{k=1}^{n}\frac{k}{(k+1)!}=\sum_{k=1}^{n}\frac{k}{P_{k+1}}=\sum_{k=1}^{n}(\frac1{P_k}-\frac1{P_{k+1}})=\frac1{P_1}-\frac1{P_2}+\frac1{P_2}-\frac1{P_3}+...+\frac1{P_n}-\frac1{P_{n+1}}=\frac1{P_1}-\frac1{P_{n+1}}=\frac1{1!}-\frac1{(n+1)!}=1-\frac1{(n+1)!}

    b)Din pacate editorul de ecuatii nu afiseaza corect(deloc) ecuatiile pentru aceasta problema. Voi incarca maine o poza cu rezolvarea.

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  4. Menim

    Menim

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    11.a) \sum_{k=1}^n(k-1)!\cdot k^2=\sum_{k=1}^n(k-1)!\cdot k\cdot k=\sum_{k=1}^nk!\cdot k=\sum_{k=1}^nk!\cdot (k+1-1)=\sum_{k=1}^nk!\cdot (k+1)-\sum_{k=1}^nk!=\sum_{k=1}^n(k+1)!-\sum_{k=1}^n(k)!=-(\sum_{k=1}^n(k)!-\sum_{k=1}^n(k+1)!)=-(1!-2!+2!-3!+...+n!-(n+1)!)=-(1!-(n+1)!)=-1+(n+1)!=(n+1)!-1

    b)\sum_{k=1}^nk!(k^2+k+1)=\sum_{k=1}^nk!(k^2+2k+1-k)=\sum_{k=1}^nk!((k+1)^2-k)=\sum_{k=1}^nk!(k+1)^2-\sum_{k=1}^nk!k=\sum_{k=1}^n(k+1)!(k+1)-\sum_{k=1}^nkk!=-(\sum_{k=1}^nkk!-\sum_{k=1}^n(k+1)(k+1)!)=-(1\cdot1!-2\cdot2!+2\cdot2!-3\cdot3!+...+n\cdot n!-(n+1)\cdot(n+1)!)=-(1\cdot1!-(n+1)(n+1)!)=-(1-(n+1)(n+1)!)=(n+1)(n+1)!-1

    c)\sum_{k=1}^n\frac{(k-1)!}{(k+1)!}=\sum_{k=1}^n\frac1{k(k+1)}=\sum_{k=1}^n\frac{(k+1)-k}{k(k+1)}=\sum_{k=1}^n(\frac1k-\frac1{k+1})=(\frac11-\frac12+\frac12-\frac13+...+\frac1n-\frac1{n+1})=1-\frac1{n+1}

    d)\sum_{k=1}^n\frac{(k+1)!}{(k-1)!}=\sum_{k=1}^nk(k+1)=\sum_{k=1}^nk^2+\sum_{k=1}^nk=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)(2n+1)-3n(n+1)}6=\frac{n(n+1)(2n+1-3)}6=\frac{n(n+1)(2n-2)}{6}=\frac{n(n+1)(n-1)}3

    Revenim la a 2-a suma de la exercitiul 10:
    \sum_{k=1}^n\frac{k^2+k+1}{(k+1)!k(k+1)}=\sum_{k=1}^n\frac{k^2+2k+1-k}{(k+1)!k(k+1)}=\sum_{k=1}^n\frac{(k+1)^2-k}{(k+1)!k(k+1)}=\sum_{k=1}^n\frac{(k+1)^2}{(k+1)!k(k+1)}-\sum_{k=1}^n\frac{k}{(k+1)!k(k+1)}=\sum_{k=1}^n\frac{1}{kk!}-\sum_{k=1}^n\frac1{(k+1)!(k+1)}=\frac1{1\cdot1!}-\frac1{2\cdot2!}+\frac1{2\cdot2!}-\frac1{3\cdot3!}+...+\frac1{n\cdot n!}-\frac1{(n+1)(n+1)!}=1-\frac1{(n+1)(n+1)!}

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  5. Menim

    Menim

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    Aranjamente
    1.Trebuie sa alegem 4 elemente dintr-o multime de 6, contand ordinea acestora. Numarul de moduri este deci A_6^4=\frac{6!}{(6-4)!}=\frac{6!}{2!}=\frac{720}{2}=360.

    2.Trebuie sa alegem 3 elemente dintr-o multime de 5, contand ordinea acestora. Numarul de moduri in care putem face asta este A_5^3=\frac{5!}{(5-3)!}=\frac{120}{2}=60. Numerele de 3 cifre nu pot avea insa prima cifra 0. Daca consideram prima cifra egala cu 0, atunci pentru celelalte 2 cifre trebuie sa alegem 2 elemente dintr-o multime de 4(deoarece nu il mai putem folosi pe 0). Avem A_4^2=\frac{4!}{2!}=\frac{24}{2}=12 moduri.
    Scadem din totalul de 60 cele 12 numere invalide si obtinem 60-12=48 numere.

    3.A_6^4=\frac{6!}{(6-4)!}=\frac{6!}{2!}=\frac{720}2=360
    A_6^0=\frac{6!}{(6-0)!}=1
    A_3^1=\frac{3!}{(3-1)!}=\frac{3!}{2!}=3
    A_3^2=\frac{3!}{(3-2)!}=\frac{3!}{1!}=\frac61=6
    A_7^4=\frac{7!}{(7-4)!}=\frac{7!}{3!}=\frac{5040}{6}=840
    A_7^3=\frac{7!}{(7-3)!}=\frac{7!}{4!}=\frac{5040}{24}=240
    Calculele ramase sunt adunari, scaderi etc. directe.

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  6. Menim

    Menim

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    4.a)i.\frac{A_8^6+A_8^5}{A_8^4}=\frac{\frac{8!}{(8-6)!}+\frac{8!}{(8-5)!}}{\frac{8!}{(8-4)!}}=\frac{\frac1{2!}+\frac1{3!}}{\frac1{4!}}=\frac{\frac12+\frac16}{\frac1{24}}=\frac{\frac36+\frac16}{\frac1{24}}=\frac{\frac46}{\frac1{24}}=\frac{\frac23}{\frac1{24}}=\frac23\cdot24=2\cdot8=16

    ii.\frac{A_4^3+A_5^4}{A_4^3}=\frac{A_4^3}{A_4^3}+\frac{A_5^4}{A_4^3}=1+\frac{\frac{5!}{(5-4)!}}{\frac{4!}{(4-3)!}}=1+\frac{\frac{5!}{1!}}{\frac{4!}{1!}}=1+\frac{5!}{4!}=1+5=6

    iii.\frac{A_7^4+A_6^5}{A_5^4}=\frac{\frac{7!}{(7-4)!}+\frac{6!}{(6-5)!}}{\frac{5!}{(5-4)!}}=\frac{\frac{7!}{3!}+\frac{6!}{1!}}{\frac{5!}{1!}}=\frac{4\cdot5\cdot6\cdot7+6!}{5!}=\frac{4\cdot5\cdot6\cdot7+2\cdot3\cdot4\cdot5\cdot6}{2\cdot3\cdot4\cdot5}=\frac{6\cdot7+2\cdot3\cdot6}{2\cdot3}=6+7=13

    b)i.\frac{A_7^3+A_6^4}{A_5^3}=\frac{\frac{7!}{(7-3)!}+\frac{6!}{(6-4)!}}{\frac{5!}{(5-3)!}}=\frac{\frac{7!}{4!}+\frac{6!}{2!}}{\frac{5!}{2!}}=\frac{5\cdot6\cdot7+\frac{6!}{2}}{\frac{5!}{2}}=\frac{2\cdot5\cdot6\cdot7+6!}{5!}=\frac{2\cdot5\cdot6\cdot7+2\cdot3\cdot4\cdot5\cdot6}{2\cdot3\cdot4\cdot5}=\frac{6\cdot7+3\cdot4\cdot6}{3\cdot4}=\frac{2\cdot7+4\cdot6}{4}=\frac{7+2\cdot6}{2}=\frac{19}2

    ii.\frac{A_5^4+A_6^5}{A_6^4}=\frac{\frac{5!}{(5-4)!}+\frac{6!}{(6-5)!}}{\frac{6!}{(6-4)!}}=\frac{\frac{5!}{1!}+\frac{6!}{1!}}{\frac{6!}{2!}}=\frac{5!+6!}{\frac{6!}{2}}=\frac{2(5!+6!)}{6!}=\frac{2(1+6)}{6}=\frac{7}{3}

    iii.\frac{A_7^4+A_6^5}{A_5^4}=\frac{\frac{7!}{(7-4)!}+\frac{6!}{(6-5)!}}{\frac{5!}{(5-4)!}}=\frac{\frac{7!}{3!}+\frac{6!}{1!}}{\frac{5!}{1!}}=\frac{4\cdot5\cdot6\cdot7+6!}{5!}=\frac{4\cdot5\cdot6\cdot7+2\cdot3\cdot4\cdot5\cdot6}{2\cdot3\cdot4\cdot5}=\frac{6\cdot7+2\cdot3\cdot6}{2\cdot3}=7+6=13

    c)i.\frac{A_n^{k-1}}{A_n^{k+1}+A_n^k}=\frac{\frac{n!}{(n-(k-1))!}}{\frac{n!}{(n-(k+1))!}+\frac{n!}{(n-k)!}}=\frac{\frac1{(n-k+1)!}}{\frac1{(n-k-1)!}+\frac1{(n-k)!}}=\frac{\frac1{(n-k-1)!(n-k)(n-k+1)}}{\frac1{(n-k-1)!}+\frac1{(n-k-1)!(n-k)}}=\frac{\frac1{(n-k)(n-k+1)}}{1+\frac1{n-k}}=\frac{\frac1{n-k+1}}{n-k+1}=\frac1{(n-k+1)^2}

    ii.\frac{(2n-k+2)A_{2n}^{k-3}}{A_{2n}^k+A_{2n}^{k-1}}=\frac{(2n-k+2)\frac{(2n)!}{(2n-(k-3))!}}{\frac{(2n)!}{(2n-k)!}+\frac{(2n)!}{(2n-(k-1)!)}}=\frac{(2n-k+2)\frac{(2n)!}{(2n-k+3)!}}{\frac{(2n)!}{(2n-k)!}+\frac{(2n)!}{(2n-k+1)!}}=\frac{(2n-k+2)\frac1{(2n-k+3)!}}{\frac1{(2n-k)!}+\frac1{(2n-k+1)!}}=\frac{(2n-k+2)\frac1{(2n-k)!(2n-k+1)(2n-k+2)(2n-k+3)}}{\frac1{(2n-k)!}+\frac1{(2n-k)!(2n-k+1)}}=\frac{\frac1{(2n-k+1)(2n-k+3)}}{1+\frac1{(2n-k+1)}}=\frac{\frac1{(2n-k+3)}}{2n-k+1+1}=\frac1{(2n-k+3)(2n-k+2)}

    iii.\frac{A_n^k-nA_{n-1}^{k-1}}{A_{n+1}^{k+1}}=\frac{\frac{n!}{(n-k)!}-n\frac{(n-1)!}{(n-1-(k-1))!}}{A_{n+1}^{k+1}}=\frac{\frac{n!}{(n-k)!}-\frac{n!}{(n-1-k+1)!}}{A_{n+1}^{k+1}}=\frac{\frac{n!}{(n-k)!}-\frac{n!}{(n-k)!}}{A_{n+1}^{k+1}}=\frac{0}{A_{n+1}^{k+1}}=0

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  7. Menim

    Menim

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    5.a)A_n^{n-1}=\frac{n!}{(n-(n-1))!}=\frac{n!}{(n-n+1)!}=\frac{n!}{1!}=\frac{n!}{1}=\frac{n!}{0!}=\frac{n!}{(n-n)!}=A_n^n
    b)A_{n+1}^{k+1}=\frac{(n+1)!}{(n+1-(k+1))!}=\frac{(n+1)n!}{(n+1-k-1)!}=(n+1)\frac{n!}{(n-k)!}=(n+1)A_n^k
    c)A_n^{k+1}=\frac{n!}{(n-(k+1))!}=\frac{n!}{(n-k-1)!}=(n-k)\frac{n!}{(n-k-1)!(n-k)}=(n-k)\frac{n!}{(n-k)!}=(n-k)A_n^k

    6.a)i.A_{n-5}^3=24
    \frac{(n-5)!}{(n-5-3)!}=24
    \frac{(n-5)!}{(n-8)!}=24
    (n-7)(n-6)(n-5)=24
    Pentru ca aranjamentul sa existe, este necesar ca n-5 sa fie mai mare decat 2, adica n sa fie mai mare decat 7. De aceea, fiecare dintre termenii din membrul stang sunt pozitivi si strict crescatori. Rezulta ca membrul stang este strict pozitiv, deci ecuatia are cel mult o solutie. Observam ca n=9 este solutia cautata.

    ii.A_x^4=30(x-2)(x-3)
    \frac{x!}{(x-4)!}=30(x-2)(x-3)
    (x-3)(x-2)(x-1)x=30(x-2)(x-3)
    Pentru ca aranjamentele sa existe, este necesar ca x sa fie mai mare decat 3. Rezulta ca (x-2)(x-3) este diferit de 0, deci putem imparti prin el:
    x(x-1)=30
    Cum x este mai mare decat 3, x(x-1) este strict pozitiv, deci avem cel mult o solutie. Observam ca x=6 este solutie.

    b)i.A_{n+3}^3=30\frac{P_{n+1}}{P_n}
    \frac{(n+3)!}{(n+3-3)!}=30\frac{(n+1)!}{n!}
    \frac{(n+3)!}{n!}=30(n+1)
    (n+1)(n+2)(n+3)=30(n+1)
    Pentru ca aranjamentele sa existe trebuie ca n+3 sa fie mai mare decat 2, deci n mai mare decat -1. Rezulta ca n+1 diferit de 0, deci putem imparti prin el:
    (n+2)(n+3)=30
    Deoarece n este mai mare decat -1, membrul stang este strict crescator, deci ecuatia are cel mult o solutie. Observam ca n=3 este solutia.

    ii.6(x-5)!A_{x-2}^5=x!
    6\frac{(x-2)!}{(x-2-5)!}=(x-4)(x-3)(x-2)(x-1)x
    6\frac{(x-2)!}{(x-7)!}=(x-4)(x-3)(x-2)(x-1)x
    6(x-6)(x-5)(x-4)(x-3)(x-2)=(x-4)(x-3)(x-2)(x-1)x
    Pentru ca aranjamentele sa existe, trebuie ca x-2 sa fie mai mare decat 4, adica x mai mare decat 6. Rezulta ca (x-4)(x-3)(x-2) este diferit de 0, deci putem imparti prin el:
    6(x-6)(x-5)=(x-1)x
    6(x^2-5x-6x+30)=x^2-x
    6x^2-66x+180=x^2-x
    5x^2-65x+180=0
    Impartim prin 5:
    x^2-13x+36=0
    Ecuatia de mai sus are radacinile 4 si 9. Doar 9 convine.

    c)i.\frac{A_n^{10}-A_n^8}{A_n^8}=131
    \frac{A_n^{10}}{A_n^8}-1=131
    \frac{\frac{n!}{(n-10)!}}{\frac{n!}{(n-8)!}}=132
    \frac{n!}{(n-10)!}\cdot\frac{(n-8)!}{n!}=132
    \frac{(n-8)!}{(n-10)!}=132
    (n-9)(n-8)=132
    Pentru ca aranjamentele sa existe, n trebuie sa fie mai mare decat 9, deci membrul stang este strict crescator, ecuatia avand cel mult o solutie. Observam ca n=20 este solutia cautata.

    ii.A_{n+4}^3=20A_{n+2}^1
    \frac{(n+4)!}{(n+4-3)!}=20\frac{(n+2)!}{(n+2-1)!}
    \frac{(n+4)!}{(n+1)!}=20\frac{(n+2)!}{(n+1)!}
    (n+4)!=20(n+2)!
    (n+3)(n+4)=20
    Pentru ca aranjamentele sa existe, trebuie ca n+4 sa fie mai mare decat 2, adica n mai mare decat -2. Rezulta ca membrul stang este strict crescator, deci avem cel mult o solutie. n=1 este solutia cautata.

    7.a)i.A_{2n-1}^2\leq930
    \frac{(2n-1)!}{(2n-1-2)!}\leq930
    \frac{(2n-1)!}{(2n-3)!}\leq930
    (2n-2)(2n-1)\leq930
    Pentru ca aranjamentele sa existe, este necesar ca 2n-1 sa fie mai mare decat 1, deci 2n-2 sa fie mai mare decat 0. De asemenea, daca 2n-1 este mai mare decat 1, atunci 2n este mai mare decat 2, deci n mai mare decat 1. Rezulta ca membrul stang al inegalitatii este strict crescator. Observam ca pentru n=16, obtinem 30*31=930, deci solutia inecuatiei este intervalul (1, 16].

    ii.A_x^4\leq30(x-2)(x-3)
    \frac{x!}{(x-4)!}\leq30(x-2)(x-3)
    (x-3)(x-2)(x-1)x\leq30(x-2)(x-3)
    Pentur ca aranjamentele sa existe, este necesar ca x sa fie mai mare decat 3, deci (x-3)(x-2) este strict pozitiv, deci putem imparti prin el:
    (x-1)x\leq30
    Deoarece x este mai mare decat 3, membrul stang este strict crescator. Observam ca pt x=6, membrul stang este egal cu 30, deci solutia inecuatiei este intervalul (3, 6].

    b)i.A_{n+1}^3\leq90\cdot\frac{P_{n-1}}{P_n}\cdot A_{n+1}^2
    \frac{(n+1)!}{(n+1-3)!}\leq90\cdot\frac{(n-1)!}{n!}\cdot\frac{(n+1)!}{(n+1-2)!}
    \frac{(n+1)!}{(n-2)!}\leq90\cdot\frac1{n}\cdot\frac{(n+1)!}{(n-1)!}
    (n-1)n(n+1)\leq90\cdot\frac1n\cdot n(n+1)
    (n-1)n(n+1)\leq90(n+1)
    Pentru ca aranjamentele sa existe, este necesar ca n+1 sa fie mai mare decat 2. Fiind strict pozitiv, impartim prin el:
    (n-1)n\leq90
    Deoarece n+1 este mai mare decat 2, rezulta ca n este mai mare decat 1. Membrul stang este pozitiv si strict crescator. Observam ca pt n=10, membrul stang este egal cu 90. Solutia inecuatiei este deci intervalul (1, 10].

    ii.6(x-5)!A_{x-2}^5\leq x!
    6(x-5)!\frac{(x-2)!}{(x-2-5)!}\leq x!
    6(x-5)!\frac{(x-2)!}{(x-7)!}\leq x!
    6(x-5)!(x-6)(x-5)(x-4)(x-3)(x-2)\leq x!
    6(x-6)(x-5)(x-4)(x-3)(x-2)\leq\frac{x!}{(x-5)!}
    6(x-6)(x-5)(x-4)(x-3)(x-2)\leq(x-4)(x-3)(x-2)(x-1)x
    Pentru ca aranjamentele sa existe, este necesar ca x-2 sa fie mare decat 4, deci x mai mare decat 6. Rezulta ca (x-4)(x-3)(x-2) este strict pozitiv. Impartim prin el.
    6(x-6)(x-5)\leq(x-1)x
    6(x^2-5x-6x+30)\leq x^2-x
    6x^2-66x+180\leq x^2-x
    5x^2-65x+180\leq0
    Simplificam prin 5:
    x^2-13x+36\leq0
    Radacinile polinomului din membrul stang sunt 4 si 9. Acesta este negativ pe intervalele (-inf, 4] si [9, +inf). Deoarece x trebuie sa fie mai mare decat 6, doar al doilea interval convine.

    c)i.182A_n^2\geq A_{n+2}^4
    182\frac{n!}{(n-2)!}\geq\frac{(n+2)!}{(n+2-4)!}
    182\frac{n!}{(n-2)!}\geq\frac{(n+2)!}{(n-2)!}
    182n!\geq(n+2)!
    182\geq n(n+1)
    n(n+1)\leq182
    Pentru ca aranjamentele sa existe, trebuie ca n sa fie mai mare decat 1, deci membrul stang este strict crescator. Observam ca pentru n=13, membrul stang este egal cu 182, deci solutia inecuatiei este intervalul (1, 13].

    ii.A_{10}^x\leq5A_{10}^{x-1}
    \frac{10!}{(10-x)!}\leq5\frac{10!}{(10-(x-1))!}
    \frac1{(10-x)!}\leq\frac5{(10-x+1)!}
    Inmultim cu numitorul membrului drept:
    10-x+1\leq5
    6\leq x
    Pentru ca aranjamentele sa existe, este necesar ca x sa fie mai mic sau egal cu 10, deci solutia este intervalul [6, 10].

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