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jagveerbrar
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Permutations doubt 21 Dec 2011, 08:24
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I have a question and I am solving it by two methods. I am getting two different answers and need help with the same
Question:
How many 3 digit numbers are there with at least one 7.

Sol 1:
Total without restriction = 9*10*10 = 900
Total without a '7' = 8*9*9 = 648
Therefore total with at least one '7' = 900 - 648 = 252.

Sol 2:
Single 7 can be at unit's, ten's or at hundred's place
Total numbers with 7 at Hundred's place: = 1*10*10 = 100
Total numbers with 7 at ten's place= 8*1*10 = 80
Total numbers with 7 at Unit's place= 8*10*1 = 80
So total = 100 + 80+80 = 260

Can somebody please explain where I am going wrong in the 2nd Solution?

TIA



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Permutations doubt 21 Dec 2011, 13:53
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jagveerbrar
I have a question and I am solving it by two methods. I am getting two different answers and need help with the same
Question:
How many 3 digit numbers are there with at least one 7.

Sol 1:
Total without restriction = 9*10*10 = 900
Total without a '7' = 8*9*9 = 648
Therefore total with at least one '7' = 900 - 648 = 252.

Sol 2:
Single 7 can be at unit's, ten's or at hundred's place
Total numbers with 7 at Hundred's place: = 1*10*10 = 100
Total numbers with 7 at ten's place= 8*1*10 = 80
Total numbers with 7 at Unit's place= 8*10*1 = 80
So total = 100 + 80+80 = 260

Can somebody please explain where I am going wrong in the 2nd Solution?

TIA
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Your first solution is good. In your second solution, you're double-counting some possibilities. It starts out fine, but when you count the numbers with 7 in the units place, you don't want to count again the numbers which have a 7 in the tens place: you already counted those numbers in your second step. So in your last step, you only have 9 choices for the tens digit, since you don't want it to be 7. If you fix that, you'll get the right answer.

In case your method is unclear to others, what you were really counting were:

Numbers with 7 as the hundreds digit
Numbers with 7 as the tens digit and *not* as the hundreds digit
Numbers with 7 as the units digit and *not* as the tens or hundreds digit

I prefer the first approach, personally, since it's easy to mistakenly double-count things if you try to do the problem directly.
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jagveerbrar
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Permutations doubt 24 Dec 2011, 11:03
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Thank you Ian for such a detailed explanation :)

IanStewart
jagveerbrar
I have a question and I am solving it by two methods. I am getting two different answers and need help with the same
Question:
How many 3 digit numbers are there with at least one 7.

Sol 1:
Total without restriction = 9*10*10 = 900
Total without a '7' = 8*9*9 = 648
Therefore total with at least one '7' = 900 - 648 = 252.

Sol 2:
Single 7 can be at unit's, ten's or at hundred's place
Total numbers with 7 at Hundred's place: = 1*10*10 = 100
Total numbers with 7 at ten's place= 8*1*10 = 80
Total numbers with 7 at Unit's place= 8*10*1 = 80
So total = 100 + 80+80 = 260

Can somebody please explain where I am going wrong in the 2nd Solution?

TIA

Your first solution is good. In your second solution, you're double-counting some possibilities. It starts out fine, but when you count the numbers with 7 in the units place, you don't want to count again the numbers which have a 7 in the tens place: you already counted those numbers in your second step. So in your last step, you only have 9 choices for the tens digit, since you don't want it to be 7. If you fix that, you'll get the right answer.

In case your method is unclear to others, what you were really counting were:

Numbers with 7 as the hundreds digit
Numbers with 7 as the tens digit and *not* as the hundreds digit
Numbers with 7 as the units digit and *not* as the tens or hundreds digit

I prefer the first approach, personally, since it's easy to mistakenly double-count things if you try to do the problem directly.
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sandeeepsharma
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Permutations doubt 26 Dec 2011, 05:32
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you are not considering case when 7 can be on unit as well as tenths and hundreds place
you need to consider these cases also



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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