Question

How many odd natural numbers of three digits are possible if repetition of digits is allowed?

(Use combination)

Answer 1

so r Prime numbers also allowed @CO_oLBoY

Answer 2

yup all odd natural num are

Answer 3

i think its like\[\left(\begin{matrix}3 \\ 3\end{matrix}\right) + \left(\begin{matrix}5 \\ 3\end{matrix}\right) + ........ + \left(\begin{matrix}2n - 1 \\ 3\end{matrix}\right)\]
Am I correct??

Answer 4

three digit numbers start from 100
and end with 999
from 100 - 200 number of odd numbers = 50
from 200 - 300 , number of odd numbers = 50
from 300 - 400 , number of odd number = 50
from 400 - 500 , number of odd numbers = 50
from 500 - 600 , number of odd numbers = 50
from 600 - 700 , number of odd numbers = 50
from 700 - 800 , no of odd no's = 50
from 800 - 900, no of od no's = 50
and from 900 - 999 , no of odd number's = 50 ( so can u add those now)
i guess, u could use this one too :p

Answer 5

@gudden but we have to use combinition

Answer 6

u didn't mention it... @CO_oLBoY

Answer 7

sorry

Answer 8

its alright..

Answer 9

@gudden cz they dont mention what method u should use. so u've got to find out urself

Answer 10

lol... i ve got this SLMC by april 15 and the paper include combinatories... at the moment im trying to understand this subject at my top speed from a book with 645 pgs... but im only 50 pgs up ... so im not much of a use here :(

Answer 11

ok

Answer 12

no prblm

Answer 13

so @gudden wht do u think the sol. is?

Answer 14

idk something about combinations...
i have a few books realted with quantitative and qualitative reasoning...and would try to find about it in those..
so, wont u mind if i could give the answer tomorrow.. (bt am not sure , if i would get it in any of those books or not)

Answer 15

take ur time

Answer 16

my time would be around 10 hours and still no surity if i would get it or not..

Answer 17

try it http://www.algebra.com/algebra/homework/Permutations/

Answer 18

@ganeshie8 what do u think?

Answer 19

odd numbers end in 1, 3, 5, 7, 9

Answer 20

5 choices for last digit
10 choices for middle digit
9 choices for first digit

Answer 21

is my ans at the start right?

Answer 22

is this right?
\[\left(\begin{matrix}3 \\ 3\end{matrix}\right) + \left(\begin{matrix}5 \\ 3\end{matrix}\right) + ....... + \left(\begin{matrix}2n - 1 \\ 3\end{matrix}\right)\]

Answer 23

@ParthKohli got any solution?

Answer 24

ganeshie is right.

Answer 25

i think i've also mentioned we have to use combination man

Answer 26

Honestly you don't have to use combination for this one. Just as long as you realize that if you count to 1000 you had to go through 0 and 99 then you have 1000 so you can subtract off those numbers there. Then odds happen every other, so divide by 2. Just a quick sanity check of whatever answer you do end up with.

Answer 27

Since you're allowing repeats, the factorial function won't help you out @CO_oLBoY

Answer 28

its the question that demands to use combination

Answer 29

I think you're imposing a false condition on yourself based on what you think "combination" means.

Answer 30

just tell me the ans maaaaaaaaaaaaaaaaaan
is mine right???

Answer 31

If I say I have three colors of balls, Red, Blue, and Green and that I also have 3 spaces to place balls, then if I DON'T use the same color multiple times I can place them 3! ways since the first spot I have 3 choices, then the second has 2, the last has 1.

But if I am allowed to REPEAT colors then I have 3 choices for the first, 3 choices for the second, and 3 choices for the last. Completely different.

Both scenarios are combinations.

Answer 32

I don't even understand what your answer is.

Answer 33

when repition is allowed its combination. i it aint then its permutation @Kainui