Solveeit Logo
Login

Question

Question: Find the total number of \(9\) digit numbers which have all the digits different \(A)9 \times 9!\...

Find the total number of 99 digit numbers which have all the digits different
A)9×9!A)9 \times 9!
B)9!B)9!
C)10!C)10!
D)D) None of these

Explanation

Solution

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 npr{}^n{p_r}(number of arrangements) where p refers to the permutation. Also, in Combination we have r-things and among n-things are ncr{}^n{c_r} which is the number of ways.

Formula used:
The relation between the permutation and combination formula is npr=r!ncr{}^n{p_r} = r!{}^n{c_r}
Complete step-by-step solution:
Since from the given, we have found the total number of 99 digit numbers which have all the digits different
So, we need to use the single digits number only and thus we have 0,1,2,3,4,5,6,7,8,90,1,2,3,4,5,6,7,8,9 possible ways.
We will fill up the first place by all the nine-digit and not 1010 . this is because if zero comes in any nine digits and zero is included in it then it will become 88 a digit number instead of nine digits.
Therefore, the first place is filed by 99 digit only. Hence there are nine ways to fill the first digit.
Now for the rest of the digits, we use the relation npr=r!ncr{}^n{p_r} = r!{}^n{c_r} where now we have 99 remaining numbers and we need to fill 88 numbers to get to the total of nine-digit different numbers.
Hence, we have npr=r!ncr⇒9p8=8!9c8{}^n{p_r} = r!{}^n{c_r} \Rightarrow {}^9{p_8} = 8!{}^9{c_8}
Since ncr=n!r!(n−r)!{}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}} then we get 9p8=8!9c8⇒8!9!8!(9−8)!⇒9!{}^9{p_8} = 8!{}^9{c_8} \Rightarrow 8!\dfrac{{9!}}{{8!(9 - 8)!}} \Rightarrow 9!
Therefore, the total number possibility is given by 9×9!9 \times 9! ways.
Thus, the option A)9×9!A)9 \times 9! is correct.

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