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How many numbers between 1000 and 10000 can be formed with 1,2…..9?
A.3024
B.60
C.78
D.None

seo-qna
Last updated date: 13th Jun 2024
Total views: 385.2k
Views today: 3.85k
Answer
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Hint: Here, we will use the concept of permutations to find the required answer. We will arrange any of the 4 numbers out of the total 9 numbers in that 4 digit number. We will then substitute the total numbers as 9 and the numbers to be arranged as 4 in the formula of permutations to get the required numbers which can be formed using the given numbers.

Formula Used:
We will use the following formulas:
1.${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, where $n$ is the total number of terms or elements and $r$ represents the number of elements to be arranged among them.
2.$n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$

Complete step-by-step answer:
Given numbers to be considered are: $1,2,3,4,5,6,7,8,9$
Hence, total numbers to be considered $ = 9$
Now, we are required to find how many numbers between 1000 and 10000.
As the numbers are between 1000 and 10000, we will get a four digit number.
Now, the four digits present in that number can be taken by any of the given numbers from 1 to 9.
In this question, nothing is mentioned regarding the repetition of the digits, so we will assume that repetition is not allowed.
Therefore, from the 9 numbers we have to arrange any four of them in the four digit number lying between 1000 and 10000.
Here, substituting $n = 9$ and $r = 4$ in the formula, we get,
${}^9{P_4} = \dfrac{{9!}}{{\left( {9 - 4} \right)!}} = \dfrac{{9!}}{{5!}}$
Computing the factorial using the formula $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$, we get
$ \Rightarrow {}^9{P_4} = 9 \times 8 \times 7 \times 6 = 72 \times 42 = 3024$
Therefore, with the numbers 1,2….9, we can form 3024 numbers between 1000 and 10000.
Hence, option A is the correct answer.

Note: An alternate way to solve this question is:
Since, we know that we have to find a 4 digit number and the digits which can be placed in the ones, tens, hundreds, and thousands place of that four digit number are any of the numbers 1,2,3,4…9, keeping in mind that repetition is not allowed.
Hence, we can directly say that the number of ways in which we can fill the units or ones place of that 4 digit number are 9.
This is because we have a total of 9 numbers with us.
Now, after filling the units place, we will be left with only 8 numbers.
Hence, we can fill the tens place of that four digit number by 8 ways.
Again, we will be left with any of the 7 numbers as 2 of them have already been placed.
Hence, the number of ways of filling the hundreds place will be: 7 ways.
Similarly, we can fill the last digit in 6 ways.
Now, we will multiply all the cases together such that:
The total numbers between 1000 and 10000 which can be formed with $1,2,....9$ will be :
$9 \times 8 \times 7 \times 6 = 72 \times 42 = 3024$
Therefore, we get the required answer as 3024.
Hence, there are 3024 numbers which can be formed between 1000 and 10000 using $1,2,....9$.
Hence, option A is the correct answer.