How many 5-digit numbers can be formed using $(0-9)$?

mars6svhym
2022-05-07
Answered

How many 5-digit numbers can be formed using $(0-9)$?

You can still ask an expert for help

priffEmipsy4i37m

Answered 2022-05-08
Author has **18** answers

Explanation:

You can have up to 10 combinations for each digit, times the number of... Numbers you want.

So you have $n={10}^{5}=100000$.

You can have up to 10 combinations for each digit, times the number of... Numbers you want.

So you have $n={10}^{5}=100000$.

zuzogiecwu

Answered 2022-05-09
Author has **3** answers

Step 1

The trick is to realize that a number can not start with a zero!

Now, there are $10}^{5$ ways in which the digits 0-9 can be chosen for the five places of a five digit number. Out of these, $10}^{4$ start with zero (once we start with 0, there are only 4 slots to fill, where we have 10 choices each).

Step 2

So, the number of possible five digit numbers is

${10}^{5}-{10}^{4}=9\times {10}^{4}=90000$

These are the numbers 10000 to 99999. We could, of course, just have counted them to get $99999-10000+1=90000$.

The trick is to realize that a number can not start with a zero!

Now, there are $10}^{5$ ways in which the digits 0-9 can be chosen for the five places of a five digit number. Out of these, $10}^{4$ start with zero (once we start with 0, there are only 4 slots to fill, where we have 10 choices each).

Step 2

So, the number of possible five digit numbers is

${10}^{5}-{10}^{4}=9\times {10}^{4}=90000$

These are the numbers 10000 to 99999. We could, of course, just have counted them to get $99999-10000+1=90000$.

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