Question

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

Answer 1

Basic number system to create any digit number will be used. Also, we will calculate the number of ways in which the given digits can be arranged to obtain different digits numbers.

Complete step-by-step answer:
Total number of positive integers formed using any number of digits from $0,1,2,3,4,5$ without repetition any digit, can be calculated by calculating one digit, two digits, three digits, four digits, five digits and six digits numbers and at last we will add all of them.
Total one digit numbers that can be formed using $0,1,2,3,4,5$ is $5$, since $0$ is not a positive integer therefore it is not considered.
Total number of two digits number we will check how many digits can be place at ten’s place i.e. except $0$ these $1,2,3,4,5$ numbers can be placed, one number among $0,1,2,3,4,5$ is selected at ten’s place therefore remaining numbers left are $5$. So, the total number of two digits numbers will be $5 \times 5 = 25$.
Similarly, the total number of three digit numbers are $5 \times 5 \times 4 = 100$, four digit numbers that can be formed using $0,1,2,3,4,5$ are $5 \times 5 \times 4 \times 3 = 300$, total number of five digit numbers are $5 \times 5 \times 4 \times 3 \times 2 = 600$ and the total number of six digits numbers are $5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600$.
Adding the total number of one digit, two digits, three digits, four digits, five digits and six digits numbers, $5 + 25 + 100 + 300 + 600 + 600 = 1630$.
It is obvious that numbers greater than $3000$ will be five digits, six digits and some four digits numbers. Now, we shall calculate the total number of four digit numbers that are greater than $3000$.
We have only three such digits that can be fixed at thousand’s place i.e. $3,4,5$, therefore numbers left to choose for hundreds place are $5$, numbers left to be fixed at ten’s place are $4$ and remaining numbers left to be fixed at one’s place are$3$. So, the total number of four digit numbers greater than $3000$ will be $3 \times 5 \times 4 \times 3 = 180$.
Adding five digits, six digits and four digit numbers greater than $3000$,
$600 + 600 + 180 = 1380$.
Therefore, the total number of positive integers which can be formed by using any number of digits from $0,1,2,3,4,5$ but using each digit not more than once in each number are $1630$ and total number of positive integers that are greater than $3000$ are $1380$.

1380 positive integers can be formed which are greater than 3000.

Note: Permutations deals with arrangement and combination deals with selection. Order is important in permutations and order is not important in Combinations. In such types of problems, we must check for the condition i.e. whether repetition of digits is allowed or not and solve the question keeping in consideration the given condition. For example, if in the above question we were given that repetition of digits was allowed then a single digit could have been used any number of times.