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by DanaJ » Mon Jan 26, 2009 7:34 am
This is a tricky one, IMHO.
The number of numbers (I know, sounds crappy... ) made up with those four digits is 4^4 = 16^2 = 256 (four digits and four positions). Now we have to subtract the number of numbers (grrr.... ) that are greater than 4321. Well, first of all, 1 or 2 cannot be the hundreds digits.
Now suppose we're looking for numbers with the thousands digit 4 and the hundreds digit either 4 or 3 (meaning smth like 43xx and 44xx). The tens digit and the units digit can be either 1,2,3 or 4. For each of the two possible "formats" (43xx and 44xx) we have 2^4 (two spots to fill and four "candidates"), so this is in total 16*2 = 32 possible numbers. We must not however, forget about the numbers that have a 43xx format but are smaller or equal to 4321. These are: 4311, 4312, 4313, 4314, 4321 or 5 numbers. This means that from our initial 256 we need to subtract (32 - 5) = 27 numbers.
So our final result will be 256 - 27 = 229.
Took me 10 mins... I promise I will say screw it and guess it if this happens on the real thing!
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Re: PNC question

by Brent@GMATPrepNow » Mon Jan 26, 2009 7:44 am
apoorva.srivastva wrote:How many natural numbers not exceeding 4321 can be formed with digits 1,2,3,4 if digits can repeat?

the answer is 229.

kindly help with the solution.

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apoorva
Are you certain that the answer is 229, apoorva?
I agree with Dana that there are 4^4 (256) four-digit numbers using the digits 1, 2, 3, and 4. And I agree that there are 27 four-digit numbers greater than 4321. This gives us 256-27=229 four-digit numbers that satisfy the conditions.

However, we must also include one-digit numbers (4 of them), two-digit numbers (4^2 of them), and three-digit numbers (4^3 of them).
So, we must add 4 + 4^2 + 4^3 = 84 to our previous total of 229 to get 313
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by gaggleofgirls » Mon Jan 26, 2009 8:01 am
Clearly, my brain has not yet kicked into gear this morning.

I did the number of 4-digit numbers to 4444.

ugh.

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by apoorva.srivastva » Mon Jan 26, 2009 8:07 am
but have u considered 1 digit ,two digit and three digit nos. while solving

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by DanaJ » Mon Jan 26, 2009 8:11 am
I'm sorry... I missunderstood that the question was about 4-digit numbers...
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by gaggleofgirls » Mon Jan 26, 2009 8:16 am
since it asked for natural numbers using the digits 1,2,3,4, doesn't it have to be 4 digit numbers only since 3 digit, 2 digit and 1 digit numbers have 0 in the 1st, 1st and 2nd or 1st, 2nd and 3rd digit?

Or at least that is one way that you could look at it to eliminate the 1 , 2 anc 3-digit numbers.

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Re: PNC question

by apoorva.srivastva » Mon Jan 26, 2009 8:16 am
well brent the answer given in the book is wrong as 229......................
the correct answer 313 as u have rightly pointed out .
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by Brent@GMATPrepNow » Mon Jan 26, 2009 9:23 am
gaggleofgirls wrote:since it asked for natural numbers using the digits 1,2,3,4, doesn't it have to be 4 digit numbers only since 3 digit, 2 digit and 1 digit numbers have 0 in the 1st, 1st and 2nd or 1st, 2nd and 3rd digit?

Or at least that is one way that you could look at it to eliminate the 1 , 2 anc 3-digit numbers.

-Carrie
If we say that a 3-digit number such as 234 has a zero in the thousands position (i.e., 0234) and this makes it ineligible to be counted, then we also have to recognize that any 4-digit number such as 2321 will also be ineligible since it has zeros preceding it as well (i.e., 02321). This assumption would make the answer to the original question zero since all numbers can be viewed the same way.
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by gaggleofgirls » Mon Jan 26, 2009 9:32 am
Brent Hanneson wrote:
If we say that a 3-digit number such as 234 has a zero in the thousands position (i.e., 0234) and this makes it ineligible to be counted, then we also have to recognize that any 4-digit number such as 2321 will also be ineligible since it has zeros preceding it as well (i.e., 02321). This assumption would make the answer to the original question zero since all numbers can be viewed the same way.
I don't agree.

Then you can just read the question as what infinite number of digit number can be created where digits 5-infinity are 0 and digits 1-4 are comprised of the digits 1,2,3,4 with repeats.

That will get you the 229 answer.

But I see what you are saying. The phrase "natural numbers" is a bit ambiguous here.

The question would be better with How many 4-digit numbers can be made...

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by krisraam » Tue Jan 27, 2009 5:13 am
The Answer is 313.


Using the digits 1,2,3,4 you can form 123 but not 0123. Because 0 is not in the set of values 1,2,3,4.

Total no of numbers = 4 + 4^2 + 4^3 + 4^4 = 340

No. greater than 4321 = 16(44--) +7(43--) + 3(432_) + 1 = 27

So answer is 313
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