Bonjour !
1.
[tex]\left(\begin{array}{ccc}n\\3\\\end{array}\right)=n\\\\\\<=>\frac{n!}{3!(n-3)!} =n\\\\<=>\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!} =n\\\\<=>\frac{n(n-1)(n-2)}{3!} =n\\\\<=>\frac{(n^2-n)(n-2)}{6} =n\\\\<=>\frac{n^3-2n^2-n^2+2n}{6} =n\\\\<=>\frac{n^3-3n^2+2n}{6} =n\\\\<=>n^3-3n^2+2n=6n\\\\<=>n^3-3n^2+2n-6n=0\\\\<=>n^3-3n^2-4n=0\\\\<=>n(n^2-3n-4)=0\\\\<=>n(n^2+n-4n-4)=0\\\\<=>n[n(n+1)-4(n+1)]=0\\\\<=>n(n+1)(n-4)=0\\\\[/tex]
Soit :
[tex]n=0[/tex]
Soit :
[tex]n+1=0\\<=>n=-1[/tex]
Soit :
[tex]n-4=0\\<=>n=4[/tex]
Donc :
[tex]\left(\begin{array}{ccc}0\\3\\\end{array}\right)=0\\\\\\\left(\begin{array}{ccc}-1\\3\\\end{array}\right)=-1\\\\\\\left(\begin{array}{ccc}4\\3\\\end{array}\right)=4[/tex]
Donc :
[tex]\left(\begin{array}{ccc}0\\3\\\end{array}\right)=0\\\\\\\left(\begin{array}{ccc}-1\\3\\\end{array}\right)=-1\\\\\\4=4[/tex]
Pour conclure :
[tex]n\neq 0\\n\neq -1\\n=4[/tex]
La solution est donc :
[tex]n=4[/tex]
2.
[tex]\left(\begin{array}{ccc}n\\2\\\end{array}\right)=\left(\begin{array}{ccc}n\\3\\\end{array}\right)\\\\\\<=>\frac{n!}{2!(n-2)!} =\frac{n!}{3!(n-3)!}\\ \\\\<=>\frac{n(n-1)(n-2)!}{2(n-2)!} =\frac{n(n-1)(n-2)(n-3)!}{6(n-3)!} \\\\\\<=>\frac{n(n-1)}{2} =\frac{n(n-1)(n-2)}{6} \\\\\\<=>\frac{n(n-1)}{2} -\frac{n(n-1)(n-2)}{6} = 0\\\\\\<=>\frac{3n(n-1)-n(n-1)(n-2)}{6} =0\\\\\\<=>3n(n-1)-n(n-1)(n-2)=0\\\\<=>n(n-1)[3-(n-2)]=0\\\\<=>n(n-1)(3-n+2)=0\\\\<=>n(n-1)(5-n)=0\\\\[/tex]
Soit :
[tex]n=0[/tex]
Soit :
[tex]n-1=0\\<=>n=1[/tex]
Soit :
[tex]5-n=0\\<=>n=5[/tex]
Donc :
[tex]\left(\begin{array}{ccc}0\\2\\\end{array}\right) =\left(\begin{array}{ccc}0\\3\\\end{array}\right)\\\\\\\left(\begin{array}{ccc}1\\2\\\end{array}\right) =\left(\begin{array}{ccc}1\\3\\\end{array}\right)\\\\\\\left(\begin{array}{ccc}5\\2\\\end{array}\right) =\left(\begin{array}{ccc}5\\3\\\end{array}\right)\\[/tex]
Donc :
[tex]\left(\begin{array}{ccc}0\\2\\\end{array}\right) =\left(\begin{array}{ccc}0\\3\\\end{array}\right)\\\\\\\left(\begin{array}{ccc}1\\2\\\end{array}\right) =\left(\begin{array}{ccc}1\\3\\\end{array}\right)\\\\\\10=10[/tex]
Ainsi :
[tex]n\neq 0\\n\neq 1\\n=5[/tex]
La solution est donc :
[tex]n=5[/tex]
Voilà ! Je te laisse faire le dernier.
