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 Post subject: PNC-I'm lost in this question
 Post Posted: Mon Jan 26, 2009 10:13 am 
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.

regards,
apoorva


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 Post subject: Brute force
 Post Posted: Tue Jan 27, 2009 8:47 pm 
As Ron says in the class, if you cannot open the question in 15 seconds start thinking the next way to tackle it. I cannot do this PNC way. During the test I would I will try to get closer to the answer and guess.

Step 1:

ABCD is the number A can be occupied in 4 ways, B in ways, C in 4 ways, and D in 4 ways

So total is 256. If you have answer > 256 KILL IT

Step 2

1111, 2222,3333, 4444 are repeated 16 times. One set of 1111,2222,3333 is acceptable.

So total is 256 - 13 (since all combinations of 4444 are incorrect) = 243. if you have a choice > 243 - KILL IT

Step 3:

Now

444D is not possible and can be formed in 3 ways (since D can be 1,2,3). We have already taken 4444 into consideration

So total is 243 -3 = 240. So if you have answer > 240 - KILL IT

Step 4

44CD is not allowed and C and D can be filled in 3 ways each. So total is 9 ways

So total is 240-9 = 231 ways. So if you have answer > 231 - KILL IT

Step 5

Remember number greater than 4321 is not allowed.

In 43CD few are not allowed

C cannot be 3 or 4

Hence total is 231-2 =229

At this point I would guess from the remaining and move on.



So


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 Post subject: Re: PNC-I'm lost in this question
 Post Posted: Wed Jan 28, 2009 7:47 pm 
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.

regards,
apoorva


apooorva,
what is the source ?


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 Post subject:
 Post Posted: Thu Jan 29, 2009 10:08 am 
Offline
ManhattanGMAT Staff


Posts: 7146
hi.

first of all, ALWAYS post ALL the answer choices with official problems.
(if this is not an official problem, then it's in the wrong folder.)

did it say that the number has to be a four-digit number?

with the literal wording that you've given, there are more than 229 possibilities, because there are also one-, two-, and three-digit numbers.

viz.:
there are 4 one-digit numbers.
there are 4 x 4 = 16 two-digit numbers.
there are 4 x 4 x 4 = 64 three-digit numbers.

let's split the four-digit numbers into categories by their first digit:
1 _ _ _ --> 4 x 4 x 4 = 64 more numbers, since these are all less than 4321.
2 _ _ _ --> 4 x 4 x 4 = 64 more numbers, since these are all less than 4321.
3 _ _ _ --> 4 x 4 x 4 = 64 more numbers, since these are all less than 4321.

this is already 276 different numbers, and we haven't even counted the ones between 4111 and 4321 yet (which, incidentally, is the hard part of the problem).

could you please go back and examine the problem statement again?
if that's literally what it says, then this is NOT a gmatprep problem, and you should run for your life from whatever happens to be the source.


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