## open_study1 Group Title 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 2 years ago 2 years ago

1. ParthKohli Group Title

$$\color{Black}{\Rightarrow {5p1} = 5! }$$

2. ParthKohli Group Title

Oops

3. goformit100 Group Title

4*3*1

4. ParthKohli Group Title

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

5. goformit100 Group Title

|dw:1338570568852:dw|

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!! ;)

7. apoorvk Group Title

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

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

9. apoorvk Group Title

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

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

11. ParthKohli Group Title

So there is a fixed value of the last digit.

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.

And 4.

14. ParthKohli Group Title

3 * 2 * 2 = 6 * 2 = 12

15. apoorvk Group Title

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

Ooops - and 0.

17. ParthKohli Group Title

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

18. open_study1 Group Title

@apoorvk u r correct can u explain that

19. goformit100 Group Title

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

20. ParthKohli Group Title

Yeah so I am correct. 3 * 2 * 2

21. apoorvk Group Title

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

22. apoorvk Group Title

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

23. goformit100 Group Title

oooo yes i mistaken, you were write, @apoorvk

24. apoorvk Group Title

^^*odd digits

25. ParthKohli Group Title

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

26. ParthKohli Group Title

Oh ok

27. apoorvk Group Title

Lol, no problem. ;)

28. FoolForMath Group Title

It should be $$2\times 3\times 3 =12$$

29. ParthKohli Group Title

3 * 2 * 2 is the answer :/

30. apoorvk Group Title

31. ParthKohli Group Title

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

32. ParthKohli Group Title

Apoorv scroll up and see the explanation I gave

33. ParthKohli Group Title

That's what I was saying Foom :P

34. FoolForMath Group Title

Drat my typo: $$2\times 3\times 3 =18$$

35. apoorvk Group Title

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)

36. ParthKohli Group Title

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

37. apoorvk Group Title

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)

38. FoolForMath Group Title

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

39. apoorvk Group Title

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

40. ParthKohli Group Title

Lol