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open_study1

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

  • one year ago
  • one year ago

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

    • one year ago
  2. ParthKohli
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    Oops

    • one year ago
  3. goformit100
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    4*3*1

    • one year ago
  4. ParthKohli
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    \( \color{Black}{\Rightarrow 5p3 = {5! \over 2!} }\)

    • one year ago
  5. goformit100
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    |dw:1338570568852:dw|

    • one year ago
  6. LyraElizabethAdams
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    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!! ;)

    • one year ago
  7. apoorvk
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    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

    • one year ago
  8. LyraElizabethAdams
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    This would give you a total of 60 non-repeating choices.

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

    • one year ago
  10. LyraElizabethAdams
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    OH, MY GOODNESS - I just realized that myself! Thanks!

    • one year ago
  11. ParthKohli
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    So there is a fixed value of the last digit.

    • one year ago
  12. LyraElizabethAdams
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    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.

    • one year ago
  13. LyraElizabethAdams
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    And 4.

    • one year ago
  14. ParthKohli
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    3 * 2 * 2 = 6 * 2 = 12

    • one year ago
  15. apoorvk
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    @goformit100 - that last one would be '2' - won't it?|dw:1338570736125:dw|

    • one year ago
  16. LyraElizabethAdams
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    Ooops - and 0.

    • one year ago
  17. ParthKohli
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    Because there are 3 for the first, 2 for second, 2 for third, it'll be 3 * 2 * 2

    • one year ago
  18. open_study1
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    @apoorvk u r correct can u explain that

    • one year ago
  19. goformit100
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    @apoorvk BUt there is only one odd number; then last must be 2 na.

    • one year ago
  20. ParthKohli
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    Yeah so I am correct. 3 * 2 * 2

    • one year ago
  21. apoorvk
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    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

    • one year ago
  22. apoorvk
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    There are two digits according to me - '3' and '9' - if we are on earth that is. -_-

    • one year ago
  23. goformit100
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    oooo yes i mistaken, you were write, @apoorvk

    • one year ago
  24. apoorvk
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    ^^*odd digits

    • one year ago
  25. ParthKohli
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    But wait, there are 3 possibilites for the first because we don't have digits repeated. 2 for second. 2 for third.

    • one year ago
  26. ParthKohli
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    Oh ok

    • one year ago
  27. apoorvk
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    Lol, no problem. ;)

    • one year ago
  28. FoolForMath
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    It should be \(2\times 3\times 3 =12\)

    • one year ago
  29. ParthKohli
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    3 * 2 * 2 is the answer :/

    • one year ago
  30. apoorvk
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    @FoolForMath - explain please.

    • one year ago
  31. ParthKohli
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    Yes @FoolForMath they're not understanding what I'm trying to say..that's exactly what I said

    • one year ago
  32. ParthKohli
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    Apoorv scroll up and see the explanation I gave

    • one year ago
  33. ParthKohli
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    That's what I was saying Foom :P

    • one year ago
  34. FoolForMath
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    Drat my typo: \(2\times 3\times 3 =18 \)

    • one year ago
  35. apoorvk
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    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)

    • one year ago
  36. ParthKohli
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    Aren't there 3 for first, 2 for second, 2 for third because the digits cant repeat?

    • one year ago
  37. apoorvk
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    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)

    • one year ago
  38. FoolForMath
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    @apoorvk: If you put 0 in the 100th place then it won't be a 3 digit number.

    • one year ago
  39. apoorvk
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    Me dumbosta! Ahaaa.. Hmm. Right. I lose approximately 3 quarters and two nickels then. #NotPayingUp -.-

    • one year ago
  40. ParthKohli
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    Lol

    • one year ago
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