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counting [#permalink]
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03 Apr 2009, 13:25
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Guys,
If I count numbers between 11 to 99, I can find it as 9911+1 = 89. So, there are 89 numbers between 11 and 99 (both inclusive).
Is there a way to get the answer through counting methods. What I mean is that to calculate the numbers between 11 and 99, we have :
the unit digit changes from 1  9 , so a total of 9 ways I can fill the unit digit. I can also fill the tens digit in 9 ways. So, the total numbers between 11 and 99 = 9*9 = 81. This is not the correct answer. This is because I missed 20,30,40,50,60,70 and 80. How would I include these missed numbers (20,30..80) in my above calculation, so that I can find the answer just by multiplying numbers?
thanks sanjay



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Re: counting [#permalink]
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03 Apr 2009, 15:46
sanjay_gmat wrote: Guys,
If I count numbers between 11 to 99, I can find it as 9911+1 = 89. So, there are 89 numbers between 11 and 99 (both inclusive).
Is there a way to get the answer through counting methods. What I mean is that to calculate the numbers between 11 and 99, we have :
the unit digit changes from 1  9 , so a total of 9 ways I can fill the unit digit. I can also fill the tens digit in 9 ways. So, the total numbers between 11 and 99 = 9*9 = 81. This is not the correct answer. This is because I missed 20,30,40,50,60,70 and 80. How would I include these missed numbers (20,30..80) in my above calculation, so that I can find the answer just by multiplying numbers?
thanks sanjay Yes, you can use counting methods: there are 9 choices for the tens' digit (19) there are 10 choices for the units' digit (09) we therefore have 90 possible two digit numbers from 10 to 99 from 11 to 99, we need to subtract one, because we don't want to count the number 10. We thus have 89 numbers between 11 and 99 (inclusive). It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest  smallest + 1) is normally preferable.
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Re: counting [#permalink]
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03 Apr 2009, 21:27
IanStewart wrote: sanjay_gmat wrote: Guys,
If I count numbers between 11 to 99, I can find it as 9911+1 = 89. So, there are 89 numbers between 11 and 99 (both inclusive).
Is there a way to get the answer through counting methods. What I mean is that to calculate the numbers between 11 and 99, we have :
the unit digit changes from 1  9 , so a total of 9 ways I can fill the unit digit. I can also fill the tens digit in 9 ways. So, the total numbers between 11 and 99 = 9*9 = 81. This is not the correct answer. This is because I missed 20,30,40,50,60,70 and 80. How would I include these missed numbers (20,30..80) in my above calculation, so that I can find the answer just by multiplying numbers?
thanks sanjay Yes, you can use counting methods: there are 9 choices for the tens' digit (19) there are 10 choices for the units' digit (09) we therefore have 90 possible two digit numbers from 10 to 99 from 11 to 99, we need to subtract one, because we don't want to count the number 10. We thus have 89 numbers between 11 and 99 (inclusive). It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest  smallest + 1) is normally preferable. Ian, thanks for your inputs. You are right that counting method is not a preferable method in that it can involve lengthy calculations. However, y'day I came across a question on gmatclub's test m01. The question is : How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit? For this question, we need to count numbers from 324721 to 458521, which is basically the same as counting numbers from 3247 to 4586. This should be easy if one spots this pattern. However, if I just tackle the question without giving too much thought to the pattern, it's going to be a nightmare trying to find the answer. Taking a look at these different digits : hundredth digit varies from 7 to 6.  10 ways thousandth digit varies from 4 to 8.  5 ways ten thousandth digit varies from 2 to 5  14 ways hundred thousandth digit varies from 3 to 4  2 ways . total number of ways = 10*5*14*2 = 1400. However, this is not the right answer. Could you please point out the flaw in my calculations?



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Re: counting [#permalink]
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04 Apr 2009, 22:51
Tricky problem..I do the following way. Consider all numbers which are between 324,700 and 399999 with the given condition: First digit: 3 = 1 Second: 29 = 8 Third: 49 = 6 Fourth: 79 = 3 Fifth: 2 = 1 Sixth : 1 = 1 Total possibilities = 1*8*6*3*1*1 = 144
Consider all numbers which are between 400000 and 458600 with the given condition: First digit: 4 = 1 Second: 05 = 6 Third: 08 = 9 Fourth: 06 = 7 Fifth: 2 = 1 Sixth : 1 = 1 Total possibilities = 1*6*9*7*1*1 = 378
Hence ans = 144+ 378 = 522 What is the answer?



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Re: counting [#permalink]
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05 Apr 2009, 09:33
sanjay_gmat wrote: How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit? For this question, we need to count numbers from 324721 to 458521, which is basically the same as counting numbers from 3247 to 4586. This should be easy if one spots this pattern.
However, if I just tackle the question without giving too much thought to the pattern, it's going to be a nightmare trying to find the answer. Taking a look at these different digits :
hundredth digit varies from 7 to 6.  10 ways thousandth digit varies from 4 to 8.  5 ways ten thousandth digit varies from 2 to 5  14 ways hundred thousandth digit varies from 3 to 4  2 ways .
total number of ways = 10*5*14*2 = 1400. However, this is not the right answer. Could you please point out the flaw in my calculations? It's because of questions like this that I said above: "It is not straightforward, however, to adapt this method to other ranges of numbers, so the alternative approach you mentioned (largest  smallest + 1) is normally preferable." You can't easily apply the multiplication principle from counting to the problem above, without breaking the problem into several cases. The number of choices, for example, for the 'thousands' digit is not 5; it can be anything from 0 to 9, since 330,721 is, for example, a valid number here, as is 339,721. The problem here is that the number of choices you have, say, for the ten thousands' digit depends on what you choose for the hundred thousands' digit. If you choose a '3' for the hundred thousands' digit, you have eight choices for the ten thousands' digit, and if you choose a '4' for the hundred thousands', you only have six choices for the ten thousands' digit. If you really want to use the product rule to count here, you can only resolve this by considering these two cases separately. You'll then discover at each stage you have to consider different cases, and the problem becomes a bit of a mess. It's not a good approach to solving the problem. That said, you've already noticed a good method here, so I'm not sure why you're looking for another one: sanjay_gmat wrote: For this question, we need to count numbers from 324721 to 458521, which is basically the same as counting numbers from 3247 to 4586. This should be easy if one spots this pattern.
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