Question

How many numbers between 5000 and 10,000 can be formed using the digits $0,1,2,3,4,5,6,7,8,9$ each digit appearing not more than once in each number.
A.$5 \times {}^9{P_3}$
B.$5 \times {}^9{C_3}$
C.$5! \times {}^9{P_3}$
D.$5! \times {}^9{C_3}$

Answer 1

Here, we will use the concept of permutation to find the required answer. Permutation is a way of arranging a number of elements from a given set such that the order of arrangement occurs. We will first find the number of ways of arranging first position using the fact that the numbers should be $5,6,7,8,9$, because we have to find the numbers between 5000 and 10,000 . We will arrange the remaining three positions using the remaining 9 numbers. Hence, using the formula of permutations and substituting the total numbers as 9 and the numbers to be arranged as 3, we will get the required answer.

Formula Used:
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, where $n$ is the total number of terms or elements and $r$ represents the number of elements to be arranged among them

Complete step-by-step answer:
Given numbers to be considered are: $0,1,2,3,4,5,6,7,8,9$
Hence, total numbers to be considered $= 10$
Now, we are required to find the total numbers between 5000 and 10,000 using the digits $0,1,2,3,4,5,6,7,8,9$ each digit appearing not more than once in each number.
As we know, between 5000 and 10,000, 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 0 to 9, except the first digit as we have to take a number greater than 5000, thus, the first digit can be from 5 to 9.
In this question, it is mentioned that each digit appears not more than once in each number.
Hence, we will assume that repetition is not allowed.
Now, first, the number of ways of arranging 1 number out of the 5 numbers i.e. 5 to 9 in the first digit from the left $= {}^5{P_1} = \dfrac{{5!}}{{\left( {5 - 1} \right)!}} = 5$
Therefore, from the remaining 9 numbers we have to arrange any three of them in the remaining 3 digit numbers in the 4 digit number lying between 5000 and 10,000.
Here, substituting $n = 9$ and $r = 3$ in the formula ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, we get
Number of ways of arranging the remaining numbers in the remaining 3 digits $= {}^9{P_3}$
Hence, the total numbers which can be formed between 5000 and 10,000 using the digits $0,1,2,3,4,5,6,7,8,9$ each digit appearing not more than once in each number $= 5 \times {}^9{P_3}$
Therefore, option A is the correct answer.
Note: An alternate way to solve this question is:
Since, we know that we have to find 4 digit numbers and the digits which can be placed in the thousands place of those four digit numbers are any of the numbers $5,6,7,8,9$, keeping in mind that repetition is not allowed.
Hence, the number of ways of filling the thousands place $= 5$ ways
Now, the digits which can be placed in the ones, tens and hundreds place of those four digit numbers are any of the remaining 9 numbers from $0,1,2,3,4,5,6,7,8,9$, 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.
Now, we will multiply all the cases together such that:
The total numbers between 5000 and 10,000 which can be formed with $0,1,2,....9$ will be $5 \times 9 \times 8 \times 7$
This can also be written as: $5 \times {}^9{P_3}$ because, ${}^9{P_3} = \dfrac{{9!}}{{\left( {9 - 3} \right)!}} = 9 \times 8 \times 7$
Therefore, we get the required answer as $5 \times {}^9{P_3}$.
Hence, there are $5 \times {}^9{P_3}$ numbers which can be formed between 5000 and 10,000 using $0,1,2,....9$.
Hence, option A is the correct answer.