Question

For each question below, calculate the number of four-digit integers (1000 through 9999 inclusive; the _rst digit cannot be 0) that satisfy the specified

property:

1. How many four-digit integers are even (for example, 2102 and 8162, but

NOT 2001)?

2. How many four-digit integers are consisted of four distinct digits in

strictly decreasing order (for example, 9621 and 8742, but NOT 1352)? (Hint:

combination problem)

3. How many four-digit integers are consisted of two pairs of distinct digits

(for example, 1001 and 2424, but NOT 3333)?

Answer 1

1. First place of the four digit numbers can be filled in 9 ways (except 0), second and third place can be filled in 10 ways each, the fourth place can be filled in 5 ways (0, 2, 4, 6, 8).
Number of such numbers = 9x10x10x5 = 4500.

2. Let us first select four distinct digits in C(10, 4) ways = 210 ways. Now each combination of these digits can be arranged only in 1 way (descending order). Note that if 0 is selected in this four numbers it will be placed at units place. Number of such numbers = 210x1 = 210.

3. There are two pairs of distinct digits.
case 1: Suppose 0 is not among them and the two digits x and y are selected from the 9 digits in C(9, 2) = 36 ways. The digits xxyy can be permuted in 4!/(2!)(2!) = 6 ways.
Number of such numbers = 36x6 = 216.

case 2: Suppose 0 is one of the numbers and one more digit z is selcted from the 9 digits in C(9, 1) = 9 ways. The digits xx00 can be permuted in 3 ways - xx00, x0x0, x00x
Number of such numbers = 9x3 = 27.

Total numbers = 216 + 27 = 243.