Question

Find the total number of $9$ digit numbers which have all the digits different
$A)9 \times 9!$
$B)9!$
$C)10!$
$D)$ None of these

Answer 1

Hint: First, let us see the permutation combination concept.
Since the question is to find the number of ways, so we are going to use permutation and combination methods which we will study in our schools to approach the given questions to find the number of ways since the number of permutations of r-objects can be found from among n-things is ${}^n{p_r}$(number of arrangements) where p refers to the permutation. Also, in Combination we have r-things and among n-things are ${}^n{c_r}$ which is the number of ways.

Formula used:
The relation between the permutation and combination formula is ${}^n{p_r} = r!{}^n{c_r}$
Complete step-by-step solution:
Since from the given, we have found the total number of $9$ digit numbers which have all the digits different
So, we need to use the single digits number only and thus we have $0,1,2,3,4,5,6,7,8,9$ possible ways.
We will fill up the first place by all the nine-digit and not $10$ . this is because if zero comes in any nine digits and zero is included in it then it will become $8$ a digit number instead of nine digits.
Therefore, the first place is filed by $9$ digit only. Hence there are nine ways to fill the first digit.
Now for the rest of the digits, we use the relation ${}^n{p_r} = r!{}^n{c_r}$ where now we have $9$ remaining numbers and we need to fill $8$ numbers to get to the total of nine-digit different numbers.
Hence, we have ${}^n{p_r} = r!{}^n{c_r} \Rightarrow {}^9{p_8} = 8!{}^9{c_8}$
Since ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$ then we get ${}^9{p_8} = 8!{}^9{c_8} \Rightarrow 8!\dfrac{{9!}}{{8!(9 - 8)!}} \Rightarrow 9!$
Therefore, the total number possibility is given by $9 \times 9!$ ways.
Thus, the option $A)9 \times 9!$ is correct.

Note: Since the factorial can be expressed as $n! = n(n - 1)(n - 2).....2.1$ so make use of this and solved the combination and permutation method.
Since ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$ is the formula for Combination and ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$ is the formula for permutation and the only difference is the relation between the permutation and combination is ${}^n{p_r} = r!{}^n{c_r}$