Question

How many positive odd integers less than 1000 are there with all the digits distinct?

Answer 1

How many do you think?

Answer 2

I think it's 10 x 9 x 5 =450

because it says "odd" so i thought i would only consider 5 digits on the last one (1,3,5,7,9)
but i don't think it's right

Answer 3

ok, it's on the right track - though it won't guarantee that the digits are distinct

Answer 4

yeah that's where i'm confused

Answer 5

certainly the 5 digits in the last place are a good starting point

Answer 6

lets break it down into a few pieces

Answer 7

since it's numbers less than 1000, let's consider numbers with just 1 digit, then those with 2 digits, then those with 3 digits

Answer 8

so for the easy 1-digit case, how many odd integers do you have? @Chibi_Robo3

Answer 9

ok.........

first digit case: 5

Answer 10

yess

Answer 11

a total of 5 single-digit numbers
ie 1,3,5,7,9

Answer 12

so now let's move onto all numbers with 2 digits
from 10 to 99

Answer 13

would you like to go ahead and count up the number of 2-digit odd integers?
(with both digits different)

Answer 14

you could calculate that mathematically if you like, along the lines of what you did earlier

Answer 15

wait i'm trying to figure out how to do it...

will it be 9 x 5? because the first digit can't be zero but i'm not sure with the second digit. I also dont think it will consider the "distinct" part of the question

Answer 16

I'll keep track of things over here too, once you let me know what you find for the 2-digit ones

Answer 17

ok! so indeed the first digit can't be 0, so that gives you initially 9 possibilities (x 5), if you wanted all odd numbers

then lets just consider the fact that the digits need to be distinct
so for each of your 5 odd numbers, you can't have that number among the 9 first digits

eg if your 2nd digit is 3, then you can't have 3 as a first digit
so that removes 1 number from the total choices for 1st digit
ie the total choices for 1st digit would no longer be 9, but 8

Answer 18

sorry to hop in, but can't keep myself from giving this tip :

Always start considering possibilities for the digit in UNIT's place, then Ten's place, and so on...

so, for 2 digit odd numbers with distinct digits,
Firstly there are 5 possibilities in unit's place.
Then for Ten's place, 0 not possible and to make it distinct you can't use the digit you used in unit's place.

just like mira said.

Answer 19

if the first digit is even, then 5 odd digits can be paired with the first digit.. 4 x 5 =20
if the first digit is odd, then only 4 odd digits can be paired with the first digit.. 5 x 4 =20

so i think the 2nd case will have 40 ?

Answer 20

Exactly! ^

Answer 21

yes that's pretty much the approach you followed to get the 40 for the 2-digit case
Thanks for the tip @hartnn, I'll keep that in mind in future. ;)

Answer 22

thanks @hartnn

@Miracrown then next is the third case... i'll try to do it again :) then i'll tell you what i got

Answer 23

...since they just mean that for the 2-digit case, the 1st digit can't be 0

ie the 1st digit can only go from 1->9
(which gave 9 total possibilities for 1st digit
then exclude 1, since it needs to be different from the 2nd digit

so that gives 8 remaining possibilities for the 1st digit

x 5 for the 2nd digit, hence your total of 40

Answer 24

Sure, are you good so far with the 5 single-digit and 40 double-digit cases? @Chibi_Robo3

Answer 25

yes:)

Answer 26

awesome, so go right ahead!

Answer 27

i have just a quick question, is 0 zero be considered as even?

Answer 28

if you're considering 0's for the 2nd digit then, yes
since 0 is not odd
so you can count it along with the evens

Answer 29

1st digit: cant be zero
if even number: 4
2nd digit(odd): 5 2nd digit (even): 4 -consider zero
3rd digit (odd): 4 3rd digit (odd): 5
4 x 5 x 4 = 80 4 x 5 x 4= 80

if odd number: 5
2nd digit (odd): 4 2nd digit(even):5
3rd digit (odd): 3 3rd digit (odd): 4
5 x 4 x 3: 60 5 x 5 x 4= 100

so all in all, there are 320 odd number for 3rd case?

Answer 30

looks good, just a moment while I check it in detail

Answer 31

then add all the odd numbers from the 1st case to 3rd case: 5 + 40 + 320 =365!

yes finally i got it THANK YOU VERY MUCH! ^_^

Answer 32

:o excellent! well done on the 3-digit case

Answer 33

yw, anytime! :D