Question

Find the number of positive integers which can be formed by any number of digits 0, 1, 2, 3, 4 and 5 but using each digit not more than once in each number. How many of these integers are greater than 3000?

Answer 1

Zero cannot be a starting digit for any number. Therefore, while forming a 6-digit number, we can fill up the first place in 5 ways. The restriction on zero ends with the starting place. Having filled it, there are 5 figures left, and therefore, the remaining places can be filled in ⁵P₅ ways.

Therefore, the number of 6-digit numbers = 5 x ⁵P₅ = 600 Similarly,

Number of 5-digit numbers = 5 .⁵P₄ = 600

Number of 4-digit numbers = 5. ⁵P₃ = 300

Number of 3-digit numbers = 5 .⁵P₂ = 100

Number of 2-digit numbers = 5 .⁵P₂ = 25

Number of single digit numbers = 5 

Therefore, the total number of positive integers = 1630.

For finding the numbers greater than 3000, we take all the 6- and 5-digit numbers together with 4-digit numbers, starting with 3, 4 or 5. The number of 4-digit numbers starting with 3, 4 or 5 is 3 .⁵P₃ = 180. 

Therefore, the total number of integers greater than 3000 is

600 + 600 + 180 = 1380.