Question

# The number of all numbers having 5 digits, with distinct digits isA. 99999B. 9 $\times {\text{ }}{}^9{{\text{P}}_4}$C. ${}^{10}{{\text{P}}_5}$D. ${}^9{{\text{P}}_4}$

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Hint: Identify whether the given question is a permutation (i.e. if the order of the elements does not matter) or a combination (i.e. if order of elements matters). Count the number of digits available for filling the space. Carefully observe what makes a 5 digit number.

Given data
We should determine all 5 digit numbers
5 digit number = _ _ _ _ _
Each space should be filled with a distinct number
Obviously the first place cannot be taken by the number 0 (because then it becomes a 4 digit number), hence there are 9 possibilities (1-9 digits) for the first space.
If the first space is taken by digit 1_ _ _ _ then numbers available for the rest 4 spaces (0, 2, 3, 4, 5, 6, 7, 8, 9) i.e. 9 possibilities.
If the first space is taken by digit 2_ _ _ _ then numbers available for the rest 4 spaces (0, 1, 3, 4, 5, 6, 7, 8, 9) i.e. 9 possibilities.
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So on, If the first space is taken by digit 9_ _ _ _ then numbers available for the rest 4 spaces (0, 1, 2, 3, 4, 5, 6, 7, 8) i.e. 9 possibilities.
So the 4 remaining spaces need to be filled using 9 digits.
This is an unordered arrangement without repetition hence it is a permutation.
Hence ${}^9{{\text{P}}_4}$
The permutation should be performed 9 times because the first digit can be filled with 9 different distinct numbers.
Hence the number of all numbers having 5 digits, with distinct digits is 9$\times {\text{ }}{}^9{{\text{P}}_4}$. Option B is the correct answer.

Note –
In this type of questions, identifying whether the arrangement is a permutation or a combination is the key. The trick lies in recognizing the first digit cannot take the value 0 and it is to be chosen from the remaining 9 digits. Followed by choosing the rest of the four spaces from 9 possibilities every time a number takes the first position.