Question

Calculati 2C₄²-3A₄¹.

Va rog mult sa ma ajutati este urgent

Answer 1

$C_{4}^{2} = \frac{4!}{2!*(4-2)!}= \frac{2!*3*4}{2!*2!}= \frac{12}{2}=6$
$A_{4}^{1}= \frac{4!}{(4-1)!}= \frac{4!}{3!}= \frac{3!*4}{3!}=4$
2*6-3*4=12-12=0