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Numbers with at least One 7

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Numbers with at least One 7

by Bullzi » Thu Sep 24, 2015 8:32 am
How many positive 3 digit integers contain at least one 7

While I understand the Restrictive Rule way of solving this question below, I couldn't arrive at the same solution using a straightforward method. Please help!

Restrictive Rule - Ignore the problem restrictions and arrive at
a) 900 ways of forming any 3 digit number (9*10*10 ways for 3 digits)
b) 648 ways of forming a 3 digit number with no 7s (8*9*9 ways for digits)
c) 900 - 648 resulting in 252

This is how I approached the problem and I am obviously missing something,:shock:
Straightforward
a) 3 digits with One 7 - 8*10*1
b) 3 digits with Two 7 - 8*1*1
c) 3 digits with Three 7 - 1*1*1


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Bullzi
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by [email protected] » Thu Sep 24, 2015 8:43 am
Hi Bullzi,

I'm going to give you some 'hints' so that you can continue to work on this problem:

The math that you use in your approach is meant to break the total number of integers down into smaller groups (which is fine). However, you have to think about what each calculation actually 'translates' into...

For example, if you're looking for the number of 3-digit numbers with TWO 7s, does it matter WHICH TWO digits are 7s?

In the calculation (8)(1)(1)....you assume that the LAST TWO digits have to be the 7s. However, there are OTHER ways to have two 7s that this calculation does not account for.... Couldn't the first and second digits be the 7s? Or the first and third digits? How can you account for those possibilities?

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by Bullzi » Thu Sep 24, 2015 9:13 am
Thanks Rich for the hints!

So, I understand that I haven't accounted for the number of arrangements for each of these 3 digit combinations I have arrived at. For example, I have assumed in a that the lone 7 is at the units position while practically it could be at any of the remaining positions too

So, would that mean that I could arrange the numbers themselves in 3! ways resulting in 8*10*1 * 3! = 480 for the option a?
I still see that my numbers are way off the expected solution..

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by Bullzi » Thu Sep 24, 2015 9:35 am
Alright, another try at getting it all right..
I figured that b and c are 8+3 = 11 (the two 7s and the third digit can be arranged 3 different ways) and 1
That leaves a as 80 * 3, however, I am unable to explain myself why! Why isn't a 80 * 3!

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by Matt@VeritasPrep » Thu Sep 24, 2015 9:56 am
Bullzi wrote:Alright, another try at getting it all right..
I figured that b and c are 8+3 = 11 (the two 7s and the third digit can be arranged 3 different ways) and 1
That leaves a as 80 * 3, however, I am unable to explain myself why! Why isn't a 80 * 3!

Thanks
Bullzi
One trick to most counting questions is how to correct for overcounting.

For instance, suppose I just count all the two digit numbers with a 7 in them. If I only consider the units place, I have 17, 27, 37, 47, 57, 67, 77, 87, 97. If I only consider the tens place, I have 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79.

So there should be 10 + 9 such numbers, right? But we've OVERCOUNTED 77: it appeared on both lists, yet we only want to count it once.
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by jervizeloy » Tue Aug 09, 2016 9:52 am
Matt@VeritasPrep wrote:
Bullzi wrote:Alright, another try at getting it all right..
I figured that b and c are 8+3 = 11 (the two 7s and the third digit can be arranged 3 different ways) and 1
That leaves a as 80 * 3, however, I am unable to explain myself why! Why isn't a 80 * 3!

Thanks
Bullzi
One trick to most counting questions is how to correct for overcounting.

For instance, suppose I just count all the two digit numbers with a 7 in them. If I only consider the units place, I have 17, 27, 37, 47, 57, 67, 77, 87, 97. If I only consider the tens place, I have 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79.

So there should be 10 + 9 such numbers, right? But we've OVERCOUNTED 77: it appeared on both lists, yet we only want to count it once.
Guys, just to make sure I have clear concepts I'm solving this problem considering the restrictions. For doing so, I'm calculating 3-digit numbers with 1 seven, 2 sevens and 3 sevens. However, I'm not getting the same result. How would you guys do it?
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by [email protected] » Tue Aug 09, 2016 7:52 pm
Hi jervizeloy,

Since you have not shown any work, it's unclear what mistake(s) you might be making. As a basis for comparison though, I'm going to start the calculation off for you, and you can finish it.

3-digit numbers with JUST one 7...

First digit is the 7: (1)(9)(9).... since the last two digits can be 0, 1, 2, 3, 4, 5, 6, 8, or 9.... but not 7.
Second digit is the 7: (8)(1)(9).... the sine first digit cannot be 0 or 7
Third digit is the 7: (8)(9)(1)

Now, how would the calculations appear if TWO of the digits were 7s?

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