anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ParthKohli
  • ParthKohli
\( \color{Black}{\Rightarrow {5p1} = 5! }\)
ParthKohli
  • ParthKohli
Oops
goformit100
  • goformit100
4*3*1

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ParthKohli
  • ParthKohli
\( \color{Black}{\Rightarrow 5p3 = {5! \over 2!} }\)
goformit100
  • goformit100
|dw:1338570568852:dw|
anonymous
  • 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
  • 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
  • anonymous
This would give you a total of 60 non-repeating choices.
apoorvk
  • apoorvk
FOLKS! @ParthKohli and @LyraElizabethAdams - seems like you missed out that 'odd' word in the question - just like I did in the beginning.
anonymous
  • anonymous
OH, MY GOODNESS - I just realized that myself! Thanks!
ParthKohli
  • ParthKohli
So there is a fixed value of the last digit.
anonymous
  • 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
  • anonymous
And 4.
ParthKohli
  • ParthKohli
3 * 2 * 2 = 6 * 2 = 12
apoorvk
  • apoorvk
@goformit100 - that last one would be '2' - won't it?|dw:1338570736125:dw|
anonymous
  • anonymous
Ooops - and 0.
ParthKohli
  • ParthKohli
Because there are 3 for the first, 2 for second, 2 for third, it'll be 3 * 2 * 2
anonymous
  • anonymous
@apoorvk u r correct can u explain that
goformit100
  • goformit100
@apoorvk BUt there is only one odd number; then last must be 2 na.
ParthKohli
  • ParthKohli
Yeah so I am correct. 3 * 2 * 2
apoorvk
  • 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
  • apoorvk
There are two digits according to me - '3' and '9' - if we are on earth that is. -_-
goformit100
  • goformit100
oooo yes i mistaken, you were write, @apoorvk
apoorvk
  • apoorvk
^^*odd digits
ParthKohli
  • ParthKohli
But wait, there are 3 possibilites for the first because we don't have digits repeated. 2 for second. 2 for third.
ParthKohli
  • ParthKohli
Oh ok
apoorvk
  • apoorvk
Lol, no problem. ;)
anonymous
  • anonymous
It should be \(2\times 3\times 3 =12\)
ParthKohli
  • ParthKohli
3 * 2 * 2 is the answer :/
apoorvk
  • apoorvk
@FoolForMath - explain please.
ParthKohli
  • ParthKohli
Yes @FoolForMath they're not understanding what I'm trying to say..that's exactly what I said
ParthKohli
  • ParthKohli
Apoorv scroll up and see the explanation I gave
ParthKohli
  • ParthKohli
That's what I was saying Foom :P
anonymous
  • anonymous
Drat my typo: \(2\times 3\times 3 =18 \)
apoorvk
  • 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
  • ParthKohli
Aren't there 3 for first, 2 for second, 2 for third because the digits cant repeat?
apoorvk
  • 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
  • anonymous
@apoorvk: If you put 0 in the 100th place then it won't be a 3 digit number.
apoorvk
  • apoorvk
Me dumbosta! Ahaaa.. Hmm. Right. I lose approximately 3 quarters and two nickels then. #NotPayingUp -.-
ParthKohli
  • ParthKohli
Lol

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