Question

(FURG-RS) Determine o valor de n
que satisfaz a equação: An-1, 3 : An, 3 = 3/4

a) 4
b) 5
c) 11
d) 12
e) 13

No gabarito aparece a alternativa C como correta, mas a minha conta bateu com a letra B. Vejam:

(n-1).(n-2).(n-3)
----------------- = 3/4
n.(n-1).(n-2)

(n-3) : n = 3/ 4
4.(n-3) : n = 3
4n - 12 -3 : n = 0
4n -15 = n
4n - n - 15 = 0
3n - 15 = 0
3n = 15
n = 15/3
n = 5

Answer 1

A parte dos arranjos está certa, vou pular essa então

$\boxed{\boxed{A_{(n-1),3}\div A_{n,3}=\dfrac{3}{4}}}\\\\\\\dfrac{A_{(n-1),3}}{A_{n,3}}=\dfrac{3}{4}\\\\\\\dfrac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)}=\dfrac{3}{4}\\\\\\\dfrac{n-3}{n}=\dfrac{3}{4}$

Multiplicando em cruz:

$4(n-3)=3n\\4n-12=3n\\4n-3n=12\\n=12$

Answer 2

$\dfrac{A_{n-1,3}}{A_{n,3}}=\dfrac{3}{4}$

Veja que:

$A_{n-1,3}=\dfrac{(n-1)!}{(n-1-3)!}=\dfrac{(n-1)!}{(n-4)!}=(n-1)(n-2)(n-3)$

$A_{n,3}=\dfrac{n!}{(n-3)!}=\dfrac{n(n-1)(n-2)(n-3)!}{(n-3)!}=n(n-1)(n-2)$

Assim:

$\dfrac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)}=\dfrac{3}{4}$

$\dfrac{n-3}{n}=\dfrac{3}{4}$

$4(n-3)=3n$

$4n-12=3n$

$n=12$

De fato, pois:

$A_{n-1,3}=A_{11,3}=\dfrac{11!}{8!}=11\cdot10\cdot9$

$A_{n,3}=A_{12,3}=\dfrac{12!}{9!}=12\cdot11\cdot10$

$\dfrac{11\cdot10\cdot9}{12\cdot11\cdot10}=\dfrac{9}{12}=\dfrac{3}{4}$

Alternativa D é a correta ^^