tHE NUMBER OF WAYS 3 DIGIT NUMBER odd number CAN BE FORMED BY 0,3,4,8,9 SO THAT THE NUMBER DOES NOT REPEAT

- anonymous

- schrodinger

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- ParthKohli

\( \color{Black}{\Rightarrow {5p1} = 5! }\)

- ParthKohli

Oops

- goformit100

4*3*1

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## More answers

- ParthKohli

\( \color{Black}{\Rightarrow 5p3 = {5! \over 2!} }\)

- goformit100

|dw:1338570568852:dw|

- anonymous

Well, because you're looking for ways in which a number does not repeat, you're going to need to use permutation. So, you'd take the total numbers that you have (5), and "permut" that by the 3 choices of number that you need. So: 5P3 Hope this helps!! ;)

- apoorvk

odd no. - so the last digit can be either 9 or 3 -> 2 ways.
choose and arrange any 2 outta remaining 4 digits = 4P2
so no. of possible of possible nos. = 2 * 4P2

- anonymous

This would give you a total of 60 non-repeating choices.

- apoorvk

FOLKS! @ParthKohli and @LyraElizabethAdams - seems like you missed out that 'odd' word in the question - just like I did in the beginning.

- anonymous

OH, MY GOODNESS - I just realized that myself! Thanks!

- ParthKohli

So there is a fixed value of the last digit.

- anonymous

You'd need to make sure, then, that you only count the numbers for which 8 is not at the end - 8 is an even number and would make the number even.

- anonymous

And 4.

- ParthKohli

3 * 2 * 2 = 6 * 2 = 12

- apoorvk

@goformit100 - that last one would be '2' - won't it?|dw:1338570736125:dw|

- anonymous

Ooops - and 0.

- ParthKohli

Because there are 3 for the first, 2 for second, 2 for third, it'll be 3 * 2 * 2

- anonymous

@apoorvk u r correct can u explain that

- goformit100

@apoorvk BUt there is only one odd number; then last must be 2 na.

- ParthKohli

Yeah so I am correct.
3 * 2 * 2

- apoorvk

I just told you - the last digit has to be odd for the whole no. to be odd. and you have two odd digits in the array - 3 and 9. so choosing any one of them = 2 ways.
now the remaining two digits = i can use any 2 from the 4 digits now left in the array, so 4P2, or 4x3.
so, ways of selecting the no. = 2 x (4x3) = 24

- apoorvk

There are two digits according to me - '3' and '9' - if we are on earth that is. -_-

- goformit100

oooo yes i mistaken, you were write, @apoorvk

- apoorvk

^^*odd digits

- ParthKohli

But wait, there are 3 possibilites for the first because we don't have digits repeated.
2 for second.
2 for third.

- ParthKohli

Oh ok

- apoorvk

Lol, no problem. ;)

- anonymous

It should be \(2\times 3\times 3 =12\)

- ParthKohli

3 * 2 * 2 is the answer :/

- apoorvk

@FoolForMath - explain please.

- ParthKohli

Yes @FoolForMath they're not understanding what I'm trying to say..that's exactly what I said

- ParthKohli

Apoorv scroll up and see the explanation I gave

- ParthKohli

That's what I was saying Foom :P

- anonymous

Drat my typo:
\(2\times 3\times 3 =18 \)

- apoorvk

After a lot of days, I guess I can bet my bottom dollar on this. -_- ( lol I barely have any USD left with me right now to lose :P)

- ParthKohli

Aren't there 3 for first, 2 for second, 2 for third because the digits cant repeat?

- apoorvk

Hmm. okay - howzzit '3' choices for the second last digit - after I have chosen one od one, I still have 4 choices in the bank - I can pick any outta those 4. so that makes or 4 choices for the middle digit (or the first one, doesn't really matter)

- anonymous

@apoorvk: If you put 0 in the 100th place then it won't be a 3 digit number.

- apoorvk

Me dumbosta!
Ahaaa.. Hmm. Right. I lose approximately 3 quarters and two nickels then. #NotPayingUp -.-

- ParthKohli

Lol

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