How many $11$-digit numbers are there which contain all of the digits $0$-$9$ at least once?
Note that in this question a number cannot begin with a digit of $0$.
My instructor explained this as a case problem, but rushed through it without pausing for questions. The answer ended up being $99 \times (10!/2!)$, but I have no idea how to get there!
$\endgroup$ 14 Answers
$\begingroup$My thinking: When you have 11 items in which 9 are unique and 2 are the same numeral repeated, then that can be arranged in $\frac{11!}{2!}$ ways. Then, there are 10 different choices for which numeral is repeated. So we have $\frac{10\times11!}{2!}$. Finally, one tenth of these arrangements should have 0 as its first digit and those don't count. So we need to multiply by $\frac{9}{10}$
$$\frac{9}{10}\times\frac{10\times11!}{2!}=99\times\frac{10!}{2!}$$
$\endgroup$ $\begingroup$See for 11 numbers there can be 9 ways for each number for 1st nine places since 0 can't be the fist number and next two are repeated from 10 numbers hence 10!/2! So 99 combinations have 10!/2! Permutations hence total ways are 99*10!/2!
$\endgroup$ $\begingroup$I disagree with your formula and the other answers. Select a nonzero digit for the leftmost place and than position all $10$ digits on the remaining $10$ places. These are $$9\cdot 10!$$ possibilities. But these are only the $11$-digit numbers where the digit on the leftmost place appears a second time in the decimal representation. To get the remaining numbers, replace this digit by one of the $9$ different digits. So you get $$9\cdot9\cdot 10!$$ numbers. But each of them appears twice, because the digit that appears twice can have found its place by the initial generation of the number or by replacing the digit equal to the leading digit. So we get only$$9\cdot9\cdot 10!/2!$$additional numbers. The sum of these two quantities is$$9\cdot 10!+9\cdot9\cdot 10!/2!=495\cdot9!$$
To illustrate this process: First select a leading digit from $1,2,3,4,5,6,7,8,9$, e.g. $7$. Then select a permutation of the $10$ digits $0,1,2,3,4,5,6,7,8,9$, e.g.$2504769183$. This gives us a number were the leading digit is the digit that is contained twice:$$7\ 2504\ 7\ 69183$$Now we want to construct a number where a digit different, to the leading digit is contained twice. So select a digit different to the leading digit, this means one of $0,1,2,3,4,5,6,8,9$ There are $9$ possibilities, we select $8$. Now we replace the digit $7$, which is not on the leftmost place , by the digit $8$ and get$$7\ 2504\ 8\ 69183$$but the same number$$7\ 2504\ 8\ 691\ 8\ 3$$we get if we replace $7$ by $8$ in the number$$7\ 2504\ 8\ 691\ 7\ 3$$.
$\endgroup$ $\begingroup$10!×10×11-9!
10! is permutaion of 0,.......9
10 is the number additionally added
11 is the number of place available to add the number
9! is to eliminate numbers begin with 0
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