Question

How many ten digit numbers can be formed using \[0,1,2,3,..,9\] (repetition not allowed)?

Answer 1

Hint: The method to find the number of ten-digit numbers that can be formed by \[0,1,2,3,..,10\] without repetition meaning the numbers are made up of ten-digit places and each number can be used only once, the order is not mentioned. Hence, if the first digit is chosen then that means there are nine options for the second digit, similarly, if the second digit is chosen then there are eight options for the third digit and so on. Hence, to find all the ten ten-digits number that we can find from the numbers given in the question, we use the formula as:
\[X\times {}^{n-1}{{C}_{1}}\times {}^{n-2}{{C}_{1}}\times ...\times {}^{n-(n-1)}{{C}_{1}}\]
where \[n\] is the total number of digits asked and \[X=10\] is the value digit number we have to find as to how many of them can be formed.

Complete Step-by-step Solution
Now placing the value of \[X=10\] and \[n=10\] we get the formula as:
\[X\times {}^{n-1}{{C}_{1}}\times {}^{n-2}{{C}_{1}}\times ...\times {}^{n-(n-1)}{{C}_{1}}\]
\[=10\times {}^{10-1}{{C}_{1}}\times {}^{10-2}{{C}_{1}}\times ...\times {}^{10-(10-1)}{{C}_{1}}\]
\[=10\times {}^{9}{{C}_{1}}\times {}^{8}{{C}_{1}}\times {}^{7}{{C}_{1}}\times {}^{6}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{1}{{C}_{1}}\]
\[=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\]
\[=3628800\]

Hence, the total number of ten digits number that can be formed by ten digits without repetition is \[3628800\].

Note:
Another method to find ten digits number that can be formed by ten digits without repetition is by permutation. In permutation, there is a formula to count the total number of \[n\] digits numbers written \[r\] times without any repetition is \[P(n,r)=\dfrac{n!}{\left( n-r \right)!}\] and digit number we have to find is \[n=10\] and the digits given to form ten digit numbers is \[0,1,2,3,..,9\] which is equal to \[10\] giving the final values in the formula as:
\[P(10,10)=\dfrac{10!}{\left( 10-10 \right)!}\]
\[P(10,10)=3628800\].