calculez nr total submultimi cu 3 elemente care reprezinta cazurile favorabile :
C³₁₀=[tex] \frac{10!}{(10-3)!*3!} = \frac{10!}{7!3!} \frac{7!*8*9*10}{7!6} = \frac{720}{6}=120 [/tex]
la cazuri posibile,scriu toate submultimile de 3 elemente care il contin pe 1:
{1,2,3},{1,3,4},{1,4,5},{1,5,6},{1,6,7},{1,7,8},{1,8,9},{1,9,10} - 8 submultimi
P=caz fav/caz pos=120/8=30/2=15
