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fisa-combinatorica pls help

fisa-combinatorica

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  1. Best Answer
    maestru (V)

    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. maestru (V)

    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{2cdot5}{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!cdot8cdot9(10!-8!)}cdotfrac{7!}2=frac{9}{8cdot 9(10!-8!)}cdotfrac12=frac1{8(8!cdot9cdot10-8!)}cdotfrac12=frac1{16cdot(8!cdot90-8!)}=frac1{16cdot89cdot8!}
    frac{1!+2!+3!+4!}{5!(9!-8!)}:frac{10}{8cdot8!}=frac{1+2+6+24}{5!(9cdot8!-8!)}cdotfrac{8cdot8!}{10}=frac{33}{5!cdot8cdot8!}cdotfrac{8cdot8!}{10}=frac{33}{5!cdot10}

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

    5.Simplificam mai intai fiecare termen al multimii.
    0!=1
    2!-1!=2-1=1
    frac{5cdot4!cdot4}{2!cdot5!}=frac{5!cdot4}{2cdot5!}=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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  4. maestru (V)

    7.a)frac{P_{x+7}}{P_{x+8}}geqfrac1{16}
    frac{(x+7)!}{(x+8)!}geqfrac1{16}
    frac{(x+7)!}{(x+7)!cdot(x+8)}geqfrac1{16}
    frac1{x+8}geqfrac1{16}
    x+8leq16
    xleq8
    b)frac{(n+1)!}{(n-1)!}leq72
    frac{(n-1)!cdot ncdot (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=ncdot n!=(n+1-1)cdot n!=(n+1)n!-n!=(n+1)!-n!=P_{n+1}-P_{n}

    a)sum_{k=1}^{n} kcdot k!=sum_{k=1}^{n} kcdot P_k=sum_{k=1}^{n}(P_{k+1}-P_k)=-sum_{k=1}^{n}(P_k-P_{k+1})=-((P_1cancel{-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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  5. maestru (V)

    11.a) sum_{k=1}^n(k-1)!cdot k^2=sum_{k=1}^n(k-1)!cdot kcdot 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)!)=-(1cdot1!-2cdot2!+2cdot2!-3cdot3!+...+ncdot n!-(n+1)cdot(n+1)!)=-(1cdot1!-(n+1)(n+1)!)=-(1-(n+1)(n+1)!)=(n+1)(n+1)!-1

    c)sum_{k=1}^nfrac{(k-1)!}{(k+1)!}=sum_{k=1}^nfrac1{k(k+1)}=sum_{k=1}^nfrac{(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}^nfrac{(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}^nfrac{k^2+k+1}{(k+1)!k(k+1)}=sum_{k=1}^nfrac{k^2+2k+1-k}{(k+1)!k(k+1)}=sum_{k=1}^nfrac{(k+1)^2-k}{(k+1)!k(k+1)}=sum_{k=1}^nfrac{(k+1)^2}{(k+1)!k(k+1)}-sum_{k=1}^nfrac{k}{(k+1)!k(k+1)}=sum_{k=1}^nfrac{1}{kk!}-sum_{k=1}^nfrac1{(k+1)!(k+1)}=frac1{1cdot1!}-frac1{2cdot2!}+frac1{2cdot2!}-frac1{3cdot3!}+...+frac1{ncdot n!}-frac1{(n+1)(n+1)!}=1-frac1{(n+1)(n+1)!}

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  6. maestru (V)

    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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  8. maestru (V)

    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}}=frac23cdot24=2cdot8=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{4cdot5cdot6cdot7+6!}{5!}=frac{4cdot5cdot6cdot7+2cdot3cdot4cdot5cdot6}{2cdot3cdot4cdot5}=frac{6cdot7+2cdot3cdot6}{2cdot3}=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{5cdot6cdot7+frac{6!}{2}}{frac{5!}{2}}=frac{2cdot5cdot6cdot7+6!}{5!}=frac{2cdot5cdot6cdot7+2cdot3cdot4cdot5cdot6}{2cdot3cdot4cdot5}=frac{6cdot7+3cdot4cdot6}{3cdot4}=frac{2cdot7+4cdot6}{4}=frac{7+2cdot6}{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{4cdot5cdot6cdot7+6!}{5!}=frac{4cdot5cdot6cdot7+2cdot3cdot4cdot5cdot6}{2cdot3cdot4cdot5}=frac{6cdot7+2cdot3cdot6}{2cdot3}=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)!}-nfrac{(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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  9. maestru (V)

    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=30frac{P_{n+1}}{P_n}
    frac{(n+3)!}{(n+3-3)!}=30frac{(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!
    6frac{(x-2)!}{(x-2-5)!}=(x-4)(x-3)(x-2)(x-1)x
    6frac{(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)!}cdotfrac{(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)!}=20frac{(n+2)!}{(n+2-1)!}
    frac{(n+4)!}{(n+1)!}=20frac{(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}^2leq930
    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^4leq30(x-2)(x-3)
    frac{x!}{(x-4)!}leq30(x-2)(x-3)
    (x-3)(x-2)(x-1)xleq30(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)xleq30
    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}^3leq90cdotfrac{P_{n-1}}{P_n}cdot A_{n+1}^2
    frac{(n+1)!}{(n+1-3)!}leq90cdotfrac{(n-1)!}{n!}cdotfrac{(n+1)!}{(n+1-2)!}
    frac{(n+1)!}{(n-2)!}leq90cdotfrac1{n}cdotfrac{(n+1)!}{(n-1)!}
    (n-1)n(n+1)leq90cdotfrac1ncdot 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)nleq90
    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}^5leq 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)leqfrac{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+180leq x^2-x
    5x^2-65x+180leq0
    Simplificam prin 5:
    x^2-13x+36leq0
    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^2geq A_{n+2}^4
    182frac{n!}{(n-2)!}geqfrac{(n+2)!}{(n+2-4)!}
    182frac{n!}{(n-2)!}geqfrac{(n+2)!}{(n-2)!}
    182n!geq(n+2)!
    182geq 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}^xleq5A_{10}^{x-1}
    frac{10!}{(10-x)!}leq5frac{10!}{(10-(x-1))!}
    frac1{(10-x)!}leqfrac5{(10-x+1)!}
    Inmultim cu numitorul membrului drept:
    10-x+1leq5
    6leq 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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