Question

Calculati )va rog ....urgentttttttttttttt

Answer 1

$\it \dfrac{P_5}{C^2_5+A^2_6} = ?$

$\it C^2_5 =\dfrac{5!}{2!(5-2)!}=\dfrac{5!}{2!\cdot 3!}= \dfrac{3!\cdot4\cdot5}{2\cdot3!} =10$

$\it A^2_6= \dfrac{6!}{(6-2)!} = \dfrac{6!}{4!} =\dfrac{4!\cdot5\cdot6}{4!} =30$

$\it P_5 = 5!=1\cdot2\cdot3\cdot4\cdot5 = 120$


$\it \dfrac{P_5}{C^2_5+A^2_6} = \dfrac{120}{10+30} =\dfrac{120}{40} =3$

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[tex]\it A^2_n = 12 \Leftrightarrow \dfrac{n!}{(n-2)!} =12\Leftrightarrow \dfrac{(n-2)!\cdot(n-1)\cdot n}{(n-2)!} =12 \Leftrightarrow \\ \\\;\\ \Leftrightarrow (n-1)\cdot n=12 \\ \\ Dar , \ \ 3\ \ \cdot\ 4 =12 \ \Longrightarrow\ n=4[/tex]