Question

Dacă x=5!+3!,.............???Multumesc!!!

Answer 1

$\displaystyle x=5!+3! \Rightarrow x=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5+1 \cdot 2 \cdot 3\Rightarrow x= 120+6\Rightarrow x=126 \\ \\ y=A_5^3 \Rightarrow y= \frac{5!}{(5-3)!} \Rightarrow y= \frac{5!}{2!} \Rightarrow y= \frac{\not2! \cdot 3 \cdot 4 \cdot 5}{\not2!} \Rightarrow y=60$
$\displaystyle z=C_5^3+11 \Rightarrow z= \frac{5!}{(5-3)! \cdot 3!} +11 \Rightarrow z= \frac{5!}{2! \cdot 3!} +11\Rightarrow \\ \\ \Rightarrow z= \frac{\not3! \cdot 4 \cdot 5}{2! \cdot \not3!}+11 \Rightarrow z= \frac{20}{1 \cdot 2} +11 \Rightarrow z= \frac{20}{2} +11\Rightarrow z=10+11\Rightarrow$
$\Rightarrow z=21$

 $x=126,~y=60,~z=21 \\ \\ x=126~(cel~mai~mare~numar)$

Answer 2

x=5!+3!=5(5-1)!+3(3-1)!=5*4!+3*2!=5*24+3*2=120+6=126
y=5!/(5-3)!=5!/2!=120/2=60
z=5!/[(5-3)!*3!]+11=120/12+11=10+11=21

x>y>z => x=126 (cel mai mare numar)