Question

Alguém poderia me ajudar???

Answer 1

a) $\frac{7!}{4!} =\frac{7*6*5*4!}{4!} = 7*6*5=210$

b) $\frac{6!}{3!}=\frac{6*5*4*3!}{3!}=6*5*4=120$

c) $\frac{15!}{13!2!} =\frac{15*14*13!}{13!2!}=\frac{15*14}{2*1} =105$

d) $\frac{4!}{3!1!} =\frac{4*3!}{3!*1!}=\frac{4}{1} =4$

Answer 2

Resposta:

a)210

b)120

c)105

d)4

Explicação passo-a-passo:

a)7!÷4! = 5040÷24 = 210

b)6!÷3! = 720÷6 = 120

c)15!÷(13!x2!) = 1307674368000÷(6227020800x2) = 1307674368000 ÷ 12454041600 = 105

d)4!÷3!x1! = 24÷(6x1) = 4