Question

How many four digits numbers are there with distinct digits? Total number of arrangements of ten digits 0,1,2,3,4,5,6,7,8,9.

Answer 1

Hint: This type of question is based on the concept of Permutation, this can also be solved by logic but we are stuck by the method of logic if the problem gets somewhat complex. So we would prefer to solve them by the concept of probability. Concept states that whenever we have to find the possible arrangement then we can multiply the possible number of arrangements to get the answer.

Complete step-by-step solution:
So, going with the question as we need to make the distinct four digit numbers, with an arrangement of ten digits;
So there will be four distinct digits; that are 1st 2nd 3rd 4th
1). For $1^{st}$ place there will be 9 possible digits we can place, as total there are ten digits but we cannot place zero (0) here because placing zero at $1^{st}$ place will not make a four digit number; so there will be 9 possible distinct digits to place at first position.
2). For $2^{nd}$ place there will be again 9 distinct digits will be possible, leaving the one digits from the arrangement of ten digits we had used in $1^{st}$ place
3). Similarly, for the $3^{rd}$ place we had eight distinct digits left from the arrangement of ten digits we had, out of which one-one digit we had used at $1^{st}$ and $2^{nd}$ place.
4). Similarly going with the $4^{th}$ place we had 7 distinct digits left from the arrangement of ten digits we had initially, out of which we had used one-one each at $1^{st}$, $2^{nd}$, and at $3^{rd}$ place.
Now, according to the concept of probability we had to multiply the number of possible arrangements; as we had possible arrangements of 9 in $1^{st}$ place, then again 9 at 2nd place, then 8 at $3^{rd}$ place, then 7 at $4^{th}$ place. So total number of arrangement for four distinct digit number will be
Number of possible arrangement at $1^{st}$ place $\times$ Number of possible arrangement at $2^{nd}$ place $\times$ Number of possible arrangement at 3rd place $\times$ Number of possible arrangement at $4^{th}$ place
$9\times 9\times 8\times 7=4536$
Hence 4536 is the number of possible arrangements of four distinct digit numbers.

Note: The concept of permutation is always valid when we ask about the number of possible arrangements whether it is about numbers or words or digit. We just need to remember some basic requirements while using concepts like the use of zero digit at 1st place will have no meaning.