ABC; ABD; ABE; ABF; ACD; ACE; ACF; ADE; ADF; AEF; BCD; BCE; BCF; BDE; BDF; BEF; CDE; CDF; CEF; DEF

Anne, Bas, Claudia; Bas, Daan, Fenna

$\left(\begin{array}{l}6\\ 3\end{array}\right)=20$ ; dus $20$ keuzes ; klopt

$20$ stappen; $17$ naar rechts en $3$ omhoog (andersom kan ook); eindpunt $(\mathrm{17,3})$ (of $(\mathrm{3,17})$)

$\left(\begin{array}{c}20\\ 3\end{array}\right)=1140$ keuzes

$(\mathrm{16,4})$

$\left(\begin{array}{c}20\\ 4\end{array}\right)=4845$ keuzes

-

$\left(\begin{array}{c}12\\ 1\end{array}\right)\cdot \left(\begin{array}{c}8\\ 3\end{array}\right)=12\cdot 56=672$ keuzemogelijkheden

$(\mathrm{9,3})$ ; $\left(\begin{array}{c}12\\ 3\end{array}\right)\cdot \left(\begin{array}{c}8\\ 1\end{array}\right)=220\cdot 8=1760$ keuzes

$\left(\begin{array}{c}12\\ 2\end{array}\right)\cdot \left(\begin{array}{c}8\\ 2\end{array}\right)=66\cdot 28=1848$ keuzes

$\left(\begin{array}{c}12\\ 4\end{array}\right)=495$ keuzes

$\left(\begin{array}{c}8\\ 4\end{array}\right)=70$ keuzes

$672+1760+1848+495+70=4845$ ; Ja, klopt.

$\left(\begin{array}{c}12\\ 5\end{array}\right)=792$ volgordes

$\left(\begin{array}{c}9\\ 4\end{array}\right)=126$ volgordes

$\left(\begin{array}{c}7\\ 3\end{array}\right)=35$ manieren

$\left(\begin{array}{c}4\\ 3\end{array}\right)=4$ manieren ; $\left(\begin{array}{c}3\\ 1\end{array}\right)\cdot \left(\begin{array}{c}4\\ 2\end{array}\right)=3\cdot 6=18$ manieren

$4+18=22$ manieren

$\left(\begin{array}{c}32\\ 4\end{array}\right)=35.960$ combinaties

$(\mathrm{8,0})$ en $(\mathrm{12,4})$ ; $70$ combinaties

$(\mathrm{6,2})$ en $(\mathrm{12,4})$ ; $784$ combinaties

$(\mathrm{7,1})$, $(\mathrm{14,2})$ en $(\mathrm{21,3})$ ; $4096$ combinaties

$(\mathrm{12,4})$ ; $1820$ combinaties

$1820$ combinaties

Laat de eerste vier stappen corresponderen met de $9$’s en de volgende vier stappen met de $10$'en; de overige $24$ stappen met de rest. De route gaat dan langs $(\mathrm{2,2})$ en $(\mathrm{4,4})$, dus $36$ combinaties.

$\left(\begin{array}{c}10\\ 4\end{array}\right)=210$ viertallen

$4\cdot 3\cdot 2\cdot 1=24$ uitslagen

Vermenigvuldigen: $210\cdot 24=5040$ uitslagen

$10\cdot 9\cdot 8\cdot 7=5040$ uitslagen

-

$210$

$\frac{12!}{3!\cdot 9!}=220$