Question

cate submultimie cu nr impar de elemente are multimea{0,1,2,3,4,5,}

Answer 1

   
[tex]\displaystyle\\ n = C_6^1 + C_6^3 + C_6^5 = 6 + \frac{6\cdot 5\cdot4}{1\cdot 2\cdot3} + 6 = 6+20+6 = \boxed{32 \text{ submultimi}}[/tex]