There are 900 3-digit numbers, for we can choose the first digit 9 ways, the 2nd digit 10 ways, and the 3rd digit 10 ways. So we will subtract from 900 all the 3-digit numbers that have at least one pair of consecutive digits. Let A = the set of all 3 digit numbers with the first two digits consecutive. Let B = the set of all 3 digit numbers with the second two digits consecutive. Then we will use N(A or B) = N(A) + N(B) - N(A and B) --- N(A) = N(3-digit integers with the first two digits consecutive) 1. There are 8 ways to choose the first digit as 1 through 8, There are then 2 ways to choose the second digit consecutive with the first digit. There are then 10 ways to choose the third digit. That's 8x2x10 = 160 2. There is 1 way to choose the first digit as 9, There is then 1 way to choose the second digit consecutive with the first digit, namely to choose it as 8 There are then 10 ways to choose the third digit. That's 1x1x10 = 10 So N(A) = 160+10 = 170 ways, ------------------ N(B) = N(3-digit integers with the last two digits consecutive) 1. There are 8 ways to choose the last digit as 1 through 8, There are then 2 ways to choose the second digit consecutive with the last digit. There are then 9 ways to choose the first digit. That's 8x2x9 = 144 2. There are 2 ways to choose the last digit as 0 or 9, There is then 1 way to choose the second digit consecutive with the last digit There are then 9 ways to choose the first digit. That's 2x1x9 = 18 Or a total of 144+18 = 162 ways, ------------------------------------ N(A and B) N(3-digit integers with both the first two and the last two digits consecutive) = 1. There are 6 ways to choose the first digit 2 through 7 There are then 2 ways to choose the second digit consecutive with the first digit. There are then 2 ways to choose the third digit consecutive with the second digit. That's 6x2x2 = 24 ways. 2. There are 3 ways to choose the first digit 1, namely 101, 110, and 121 3. There are 3 ways to choose the first digit 8, namely 878, 879, and 898 4. There are 2 ways to choose the first digit 9, namely 987 and 989. That's 8 ways. So N(A and B = 24+8 = 32 ------------------------- Therefore N(A or B) = N(A) + N(B) - N(A and B) = 170 + 162 - 32 = 300, Therefore the answer to the problem is 900 - 300 or 600. Edwin