Solveeit Logo

Question

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

How many ten digit numbers can be formed using 0,1,2,3,..,90,1,2,3,..,9 (repetition not allowed)?

Explanation

Solution

The method to find the number of ten-digit numbers that can be formed by 0,1,2,3,..,100,1,2,3,..,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×n1C1×n2C1×...×n(n1)C1X\times {}^{n-1}{{C}_{1}}\times {}^{n-2}{{C}_{1}}\times ...\times {}^{n-(n-1)}{{C}_{1}}
where nn is the total number of digits asked and X=10X=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=10X=10 and n=10n=10 we get the formula as:
X×n1C1×n2C1×...×n(n1)C1X\times {}^{n-1}{{C}_{1}}\times {}^{n-2}{{C}_{1}}\times ...\times {}^{n-(n-1)}{{C}_{1}}
=10×101C1×102C1×...×10(101)C1=10\times {}^{10-1}{{C}_{1}}\times {}^{10-2}{{C}_{1}}\times ...\times {}^{10-(10-1)}{{C}_{1}}
=10×9C1×8C1×7C1×6C1×5C1×4C1×3C1×2C1×1C1=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}}
=10×9×8×7×6×5×4×3×2×1=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1
=3628800=3628800

Hence, the total number of ten digits number that can be formed by ten digits without repetition is 36288003628800.

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 nn digits numbers written rr times without any repetition is P(n,r)=n!(nr)!P(n,r)=\dfrac{n!}{\left( n-r \right)!} and digit number we have to find is n=10n=10 and the digits given to form ten digit numbers is 0,1,2,3,..,90,1,2,3,..,9 which is equal to 1010 giving the final values in the formula as:
P(10,10)=10!(1010)!P(10,10)=\dfrac{10!}{\left( 10-10 \right)!}
P(10,10)=3628800P(10,10)=3628800.