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Q. How many 3 digit odd numbers can be formed using 0,3,5,7, repetition not allowed.

WHAT I DID :-

3 x 3 x 1 = 9

For Hundredth place - It can be filled in 3 ways (any of 3,5,7), we cannot use 0.

For Tens place - It can be filled in 3 ways (from 0,3,5,7) as one of 3,5,7 already filled in hundredth place.

For Ones place - It can be filled in 1 way as two digits of 3,5,7 already used in above two places and it cannot use 0.

SOLUTION ON THE BOOK SAYS: -

It fills Hundredth first, then Ones and then Second.

solution from book

3 X 2 X 2 = 12

What is that I'm not understanding or doing wrong?? How is it determined that which order should be followed, like first we should fill hundredth placed then first place then others ?

rickenjus
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    You don't have to, but life is simpler if you do. It is often useful to take care of "restrictions" first. The middle digit is not fussy. But neither hundredth nor units digit can be $0$. – André Nicolas Oct 05 '14 at 05:55
  • Elaborate please. – rickenjus Oct 05 '14 at 05:59
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    The number must be odd, so it cannot end in $0$. Let's fix your analysis along the lines of the answer by Henry Swanson. You are right, there are $3$ ways to fill the hundreds digit. Now let's do the middle. If you use a $0$, you have $2$ choices for units digit. If you use an odd number in the middle ($2$ choices) you have only one choice for units digit. So total is $(3)[(1)(2)+(2)(1)]$. The "book" way is easier. – André Nicolas Oct 05 '14 at 06:04
  • Thanks Andre, that was helpful. – rickenjus Oct 05 '14 at 06:13
  • You are welcome. An analysis along your lines can be done, it is just that we have to break up into cases. Sometimes cases are unavoidable, but in this example leaving the choice of middle digit to the end makes cases unnecessary. – André Nicolas Oct 05 '14 at 06:18
  • Alternative method: Zero is the only digit which is constrained.

    There are two possible places for zero. Once the place for zero is chosen, there are six ways to arrange the remaining three digits. $2 \times 6 =12$

     – Mark Bennet Oct 05 '14 at 06:53

1 Answers1

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There are $3$ ways to fill the hundreds place. $3$ ways to fill the tens place. But there can be $2$ or $1$ ways to fill in the ones place. For example, $30\_$ has two options, but $35\_$ has only one. It depends on how you fill the tens place. The solution given avoids that problem because it picks the units before the tens.

Henry Swanson
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  • Thank you. I got my mistake, so is there any proper way to go through such problems like fill highest place first then come to one's placed or I have to take care of exception each digit is making. ? – rickenjus Oct 05 '14 at 06:06
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    There's no hard and fast way to do these problems. You just have to be careful that your casework is correct. Like André says, it's usually best to take care of the most restrictive parts first. This means that you have less casework to do. – Henry Swanson Oct 05 '14 at 06:09
  • Thanks Henry, will keep that in mind.

    I was earlier doubtful that, should I post my Question here or not, but man that was fast, got my answer quick.. :)

     – rickenjus Oct 05 '14 at 06:15