Question

analise combinatória fatorial de :
a)6!
b)5!+2!=
c)3!+4!-0=
d)3!-4!+1=
e)7!/3!=
f)10!/9!=
me ajudaaa rapidooo​

Answer 1

Explicação passo-a-passo:

a)6! = 6x5x4x3x2x1= 720

b)5!+2!= 5x4x3x2x1 + 2x1= 120 + 2= 122

c)3!+4!-0= 3x2x1 + 4x3x2x1= 6 + 24= 30 - 0 = 30

d)3!-4!+1= 3x2x1 - 4x3x2x1 + 1 = 6 - 24 + 1= -17

e)7!/3!= $\frac{7!}{3!} =\frac{7.6.5.4.3.2.1}{3.2.1} = \frac{5040}{6} = \frac{840}{1} = 840$  

f)10!/9!= $\frac{10!}{9!} \frac{10.9.8.7.6.5.4.3.2.1}{9.8.7.6.5.4.3.2.1} = \frac{3628800}{362880} = \frac{10}{1}$