Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum

[M7MBA]

Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?

A. 1
B. 2
C. 3
D. 4
E. 5

Answer: D

Source: GMAT Prep

[swerve]

Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?

A. 1
B. 2
C. 3
D. 4
E. 5

Answer: D

Source: GMAT Prep

From the question, we can deduce that

1) The Hundreds place will consist of \(4 + 1\)(carried forward)
2) The Units place can ONLY have \(8\) and \(7\) as no other pair of digits can form the sum with units place of '\(5\)'
3) Thus leaving us with 9 and 5 as Tens place

Now, The number of ways to arrange Units place \(= 2!\)
and the number of ways to arrange Tens place \(= 2!\)

Since both are independent events, then the total number of arrangements is \(= 2! \cdot 2! = 4\)

Therefore, D