How many positive integers less than 1000 have distinct digits?
Is there a way to approach through combinatorics?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
10 choices for the first digit, then 9 for the second and finally 8 for the third
counting principle from there
damn this could be hard..
1-digit integers: 1, ..., 9 - there are 9 integers.
2-digit integers: 10, ..., 99 - there are 90 integers, but in 9 of them (11, ..., 99) the
two digits are the same. So there are 90 − 9 = 81 2-digit integers with distinct digits.
Another way to get this answer is to consider the number of possibilities for each digit:
the first digit can be any non-zero digit, so it has 9 choices. The second digit can be any
digit except equal to the first one, so it has 9 choices too. There are 9 · 9 = 81 choices
3-digit integers: 100, ..., 999 - there are 900 integers total, but some of them have a
repeating digit... The number of 3-digit integers with distinct digits can be counted as
follows: the first digit can be any non-zero digit, so it has 9 choices. The second digit
can be any digit except equal to the first one, so it has 9 choices too. Finally, the third
digit can be any digit except equal to the first digit or the second digit, so it has 8
choices. There are 9 · 9 · 8 = 648 choices total.
So there are 9 + 81 + 648 = 738 positive integers less than 1000 wit distinct digits.
You can use some combinatorics, but really just a little multiplication.
1000 does not have unique digits. It's out
Let's look at the rest - 0 - 999
* 1-9 : 9 total, all distinct
* 10-99 : 9*9=81 distinct, because 11,22,33,44,55,66,77,88,99 are removed (that is 90-9=81 distinct numbers). Note, in the MSD, 1-9 are allowed, in the second digit, 0-9 are allow, but we need to take out the digit that was used in MSD, leaving 9 digits for the LSD
* 100-999: 9*9*8=648 distinct, for similar reasons
Total = 9+81+648 = 738 distinct numbers