Question

Determine the number of six digit odd numbers that can be created using digits 0 to 9 without repition. Describe any restrictions that exist.

Please help if you can, thanks ☼
This problem involves permutations/combinations

Answer 1

one problem though, is 012345 considered a 6 digit?

Answer 2

yes. zero is included @sourwing

Answer 3

to be an odd number, the last digit must be one, that is either 1,3,5,7,or,9

9 * 8 * 7 * 5 * 4 * 5

Answer 4

So do I use a formula or fundemental counting principle @sourwing

Answer 5

@DSS @sourwing You guys both did it differently

Answer 6

I'm not sure...you need odd numbers

Answer 7

Ok @DSS Thanks for tryng though. Regardless, it is still appreciated

Answer 8

opps i meant to say the last digit must be *odd*
and no you don't use counting principle because that's is only applied when order does not matter. But order matters in this case. So you just use permutation formula

Answer 9

Can you show me what I put in the formula please @sourwing
Like the n value
and the
r value? ☼

Answer 10

(9P5) * 5

once an odd digit is chosen for the last digit. You have pick 5 out of 9 numbers left to rearrange

Answer 11

Wouldn't it be \[10!\div(10-5)!\]
Because there are 10 options to choose from and you choose 5 of them (1,3,5,7,9)
Wouldn't there be 10 because you include zero (because it is from zero to nine)

Answer 12

@sourwing

Answer 13

no, digits CAN NOT be repeated.

Answer 14

How are they repeated? @sourwing

Answer 15

you have chose one odd number for the last digit, say 5, then there are 9 numbers left.

0,1,2,3,4,6,7,8,9 (9 numbers)

Answer 16

okay, so what restrictions would I say exist @sourwing

Answer 17

well the restriction is number can not be repeated

Answer 18

and the last number must be odd

Answer 19

okay, I think I got it now. Thank you for the help ♪

Answer 20

Oh, one more thing. Why do you multiply 5 by 9P5?? @sourwing

Answer 21

because that is the number of ways the last digit can be