Question

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Answer 1

Explicație pas cu pas:

nr pare: 0, 2, 4

nr impare: 1,3,5

x C y - combinari de x luate cate y

( 4 C 3 -1 )*3 + 3² +1=3²*2+1=19

Bafta!

Answer 2

 

$\displaystyle\bf\\Se~da~multimea~~A=\{0;~1;~2;~3;~4;~5\}\\Se~cere~numarul~submultimilor~cu~3~elementre\\dintre~care~cel~putin~un~element~sa~fie~par.\\\\Rezolvare:\\\\Avem~3~elemente~pare:~~0;~2;~4~~si~3~elemente~ipare:~~1;~3;~5.\\\\Vom~avea~submultimi~cu:\\1~element~par,~cu~2~elemente~pare~si ~cu~3~elemente~pare.$

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$\displaystyle\bf\\Numarul~de~submultimi~este:\\\\n=\underbrace{\bf C_3^1\times C_3^2}_{1\,p~si~2\,i}+\underbrace{\bf C_3^2\times C_3^1}_{2\,p~si~1\,i}+\underbrace{\bf C_3^3\times C_3^0}_{3\,p~si~0\,i}=\\\\n=~3\times3~~+~~3\times3~~+~~1\times1\\\\n=3\times3+3\times3+1\times1\\\\n=9+9+1=\boxed{\bf19~submultimi}$