BTGmoderatorDC wrote: ↑

Mon May 10, 2021 3:21 pm

How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A) 360
B) 720
C) 840
D) 1,080
E) 1,440


OA B

Source: Princeton Review

When two \(7\)s are used \((3445{\color{red}77}) = \dfrac{6!}{2!2!} = \dfrac{720}{4} =180\)

When one \(7\) and one \(8\) is used \((3445{\color{red}78}) = \dfrac{6!}{2!} = 360\)

When two \(8\)s are used \((3445{\color{red}88}) = \dfrac{6!}{2!2!} = \dfrac{720}{4} =180\)

So, the total number of ways \(= 180+360+180 = 720 \quad \Longrightarrow\quad\)B