Răspuns:
16
Explicație pas cu pas:
A={,2,3,...,50}. B⊆A, {1,2,3,...,46}⊆B
Submultimea {1,2,3,...,46} obligatoriu de contine in B.
mai sunt 4 numere ce se mai pot contine in B, anume 47,48,49,50, alaturi de cele 46.
Atunci submultimi pot fi formate 2⁴=16.
[tex]C_{4}^{0}+C_{4}^{1}+C_{4}^{2}+C_{4}^{3}+C_{4}^{4}=\frac{4!}{0!*(4-0)!}+\frac{4!}{1!*(4-1)!}+\frac{4!}{2!*(4-2)!}+\frac{4!}{3!*(4-3)!}+\frac{4!}{4!*(4-4)!}=1+4+6+4+1=16.[/tex]
