Combinatoire et dénombrement

Calculer des coefficients binomiaux avec les propriétés - Exercice 1

10 min
15
Question 1
En utilisant les propriétés du cours et sans calculatrice, déterminer :

1.\red{\bf{1.}} A=(88)A=\left(\begin{array}{c} {8} \\ {8} \end{array}\right)                                                                                                          \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2.\red{\bf{2.}} B=(131)B=\left(\begin{array}{c} {13} \\ {1} \end{array}\right)

3.\red{\bf{3.}} C=(1312)C=\left(\begin{array}{c} {13} \\ {12} \end{array}\right)                                                                                                    \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 4.\red{\bf{4.}} D=(170)D=\left(\begin{array}{c} {17} \\ {0} \end{array}\right)

Correction
  • Pour tout entier naturel nn et pour tout entier kk tel que 0kn0\le k \le n, on a : (n0)=1\left(\begin{array}{c} {n} \\ {0} \end{array}\right)=1 et (nn)=1\left(\begin{array}{c} {n} \\ {n} \end{array}\right)=1
  • Pour tout entier naturel nn non nul, on a : (n1)=n\left(\begin{array}{c} {{\color{green}{n}}} \\ {1} \end{array}\right)={\color{green}{n}}
  • Proprieˊteˊ de symeˊtrie\red{\text{Propriété de symétrie}} : Pour tout entier naturel kk compris entre 00 et n1n-1, on a : (nk)=(nnk)\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{red}{k}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{blue}{n}}-{\color{red}{k}}} \end{array}\right)
  • 1.\red{\bf{1.}} A=(88)=1A=\left(\begin{array}{c} {8} \\ {8} \end{array}\right)=1
    2.\red{\bf{2.}} B=(131)=13B=\left(\begin{array}{c} {{\color{green}{13}}} \\ {1} \end{array}\right)={\color{green}{13}}
    3.\red{\bf{3.}} C=(1312)C=\left(\begin{array}{c} {{\color{blue}{13}}} \\ {{\color{red}{12}}} \end{array}\right) ainsi : C=(131312)=(131)C=\left(\begin{array}{c} {{\color{blue}{13}}} \\ {{\color{blue}{13}}-{\color{red}{12}}} \end{array}\right)=\left(\begin{array}{c} {13} \\ {1} \end{array}\right) . Il en résulte donc que C=(1312)=(131)=13C=\left(\begin{array}{c} {{\color{blue}{13}}} \\ {{\color{red}{12}}} \end{array}\right)=\left(\begin{array}{c} {13} \\ {1} \end{array}\right)=13
    4.\red{\bf{4.}} D=(170)=1D=\left(\begin{array}{c} {17} \\ {0} \end{array}\right)=1
    Question 2

    On admet que (127)=792\left(\begin{array}{c} {12} \\ {7} \end{array}\right)=792. Calculer (125)\left(\begin{array}{c} {12} \\ {5} \end{array}\right)

    Correction
  • Proprieˊteˊ de symeˊtrie\red{\text{Propriété de symétrie}} : Pour tout entier naturel kk compris entre 00 et n1n-1, on a : (nk)=(nnk)\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{red}{k}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{blue}{n}}-{\color{red}{k}}} \end{array}\right)
  • (127)=(12127)=(125)\left(\begin{array}{c} {{\color{blue}{12}}} \\ {{\color{red}{7}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{12}}} \\ {{\color{blue}{12}}-{\color{red}{7}}} \end{array}\right)=\left(\begin{array}{c} {12} \\ {5} \end{array}\right)
    Il en résulte donc que
    (125)=792\left(\begin{array}{c} {12} \\ {5} \end{array}\right)=792
    Question 3

    On admet que (108)=45\left(\begin{array}{c} {10} \\ {8} \end{array}\right)=45 et (109)=10\left(\begin{array}{c} {10} \\ {9} \end{array}\right)=10 . Calculer (119)\left(\begin{array}{c} {11} \\ {9} \end{array}\right)

    Correction
  • Formule de Pascal \red{\text{Formule de Pascal }} :
    Pour tous entiers naturels nn et kk tels que n1n\ge 1 et 0kn10\le k \le n-1, on a : (nk)+(nk+1)=(n+1k+1)\left(\begin{array}{c} {n} \\ {k} \end{array}\right)+\left(\begin{array}{c} {n} \\ {k+1} \end{array}\right)=\left(\begin{array}{c} {n+1} \\ {k+1} \end{array}\right)
  • Nous appliquons la formule de pascal :
    (108)+(108+1)=(10+18+1)\left(\begin{array}{c} {10} \\ {8} \end{array}\right)+\left(\begin{array}{c} {10} \\ {8+1} \end{array}\right)=\left(\begin{array}{c} {10+1} \\ {8+1} \end{array}\right)
    (108)+(109)=(119)\left(\begin{array}{c} {10} \\ {8} \end{array}\right)+\left(\begin{array}{c} {10} \\ {9} \end{array}\right)=\left(\begin{array}{c} {11} \\ {9} \end{array}\right)
    45+10=(119)45+10=\left(\begin{array}{c} {11} \\ {9} \end{array}\right)
    Ainsi :
    (119)=55\left(\begin{array}{c} {11} \\ {9} \end{array}\right)=55
    Question 4

    Calculer (135134)\left(\begin{array}{c} {135} \\ {134} \end{array}\right)

    Correction
  • Proprieˊteˊ de symeˊtrie\red{\text{Propriété de symétrie}} : Pour tout entier naturel kk compris entre 00 et n1n-1, on a : (nk)=(nnk)\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{red}{k}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{blue}{n}}-{\color{red}{k}}} \end{array}\right)
  • (135134)=(135135134)=(1351)\left(\begin{array}{c} {{\color{blue}{135}}} \\ {{\color{red}{134}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{135}}} \\ {{\color{blue}{135}}-{\color{red}{134}}} \end{array}\right)=\left(\begin{array}{c} {135} \\ {1} \end{array}\right)
  • Pour tout entier naturel nn non nul, on a : (n1)=n\left(\begin{array}{c} {{\color{green}{n}}} \\ {1} \end{array}\right)={\color{green}{n}}
  • Or (1351)=135\left(\begin{array}{c} {{\color{green}{135}}} \\ {1} \end{array}\right)={\color{green}{135}}
    Il en résulte donc que
    (135134)=1\left(\begin{array}{c} {135} \\ {134} \end{array}\right)=1
    Question 5

    On admet que (126)=924\left(\begin{array}{c} {12} \\ {6} \end{array}\right)=924 et (127)=792\left(\begin{array}{c} {12} \\ {7} \end{array}\right)=792 . Calculer (137)\left(\begin{array}{c} {13} \\ {7} \end{array}\right)

    Correction
  • Formule de Pascal \red{\text{Formule de Pascal }} :
    Pour tous entiers naturels nn et kk tels que n1n\ge 1 et 0kn10\le k \le n-1, on a : (nk)+(nk+1)=(n+1k+1)\left(\begin{array}{c} {n} \\ {k} \end{array}\right)+\left(\begin{array}{c} {n} \\ {k+1} \end{array}\right)=\left(\begin{array}{c} {n+1} \\ {k+1} \end{array}\right)
  • Nous appliquons la formule de pascal :
    (126)+(126+1)=(12+16+1)\left(\begin{array}{c} {12} \\ {6} \end{array}\right)+\left(\begin{array}{c} {12} \\ {6+1} \end{array}\right)=\left(\begin{array}{c} {12+1} \\ {6+1} \end{array}\right)
    (126)+(127)=(137)\left(\begin{array}{c} {12} \\ {6} \end{array}\right)+\left(\begin{array}{c} {12} \\ {7} \end{array}\right)=\left(\begin{array}{c} {13} \\ {7} \end{array}\right)
    924+792=(137)924+792=\left(\begin{array}{c} {13} \\ {7} \end{array}\right)
    Ainsi :
    (137)=1  716\left(\begin{array}{c} {13} \\ {7} \end{array}\right)=1\;716
    Question 6
    Soit nn un entier naturel.

    Déterminer (nn1)\left(\begin{array}{c} {n} \\ {n-1} \end{array}\right)

    Correction
  • Pour tout entier naturel nn non nul, on a : (n1)=n\left(\begin{array}{c} {{\color{green}{n}}} \\ {1} \end{array}\right)={\color{green}{n}}
  • Nous savons donc que (n1)=n\left(\begin{array}{c} {{\color{green}{n}}} \\ {1} \end{array}\right)={\color{green}{n}}
  • Proprieˊteˊ de symeˊtrie\red{\text{Propriété de symétrie}} : Pour tout entier naturel kk compris entre 00 et n1n-1, on a : (nk)=(nnk)\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{red}{k}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{blue}{n}}-{\color{red}{k}}} \end{array}\right)
  • Ainsi :
    (n1)=(nn1)=n\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{red}{1}}} \end{array}\right)=\left(\begin{array}{c} {{\color{blue}{n}}} \\ {{\color{blue}{n}}-{\color{red}{1}}} \end{array}\right)=n