Question: Numbers with at Least One 7

Comment on Numbers with at Least One 7

How to solve above question

How to solve above question using strategy 1 i.e adhering the restrictions( without breaking the restrictions)

That approach would be

That approach would be incredibly time-consuming.

We would have to determine:
(1) The number of 3-digit integers with exactly one 7
(2) The number of 3-digit integers with exactly two 7's
(3) The number of 3-digit integers with exactly three 7's
Then we'd have to add all three values.

Within step (1) alone, we must keep in mind that the first digit cannot be 0 (otherwise, it's not a 3-digit integer)
So, for step (1), we must consider 3 cases:
(i) the first digit is 7
(ii) the second digit is 7, and the first digit is a non-zero digit
(iii) the third digit is 7, and the first digit is a non-zero digit

Similar steps must be taken with step (2).

The whole thing is WAYYYYYYY more complicated if we don't use the strategy explained in the video.

Let's try it.

Let's try it.

Step (1) - The number of 3-digit integers with exactly one 7
case i: 7 _ _
We can select the 2nd digit in 9 ways (any digit EXCEPT 7) and the 3rd digit in 9 ways (any digit EXCEPT 7)
Total = (9)(9) = 81

case ii: _ 7 _
We can select the 1st digit in 8 ways (any digit EXCEPT 0 and 7), and the 3rd digit in 9 ways (any digit EXCEPT 7)
Total = (8)(9) = 72

case iii: _ _ 7
We can select the 1st digit in 8 ways (any digit EXCEPT 0 and 7), and the 2nd digit in 9 ways
Total = (8)(9) = 72

So, total for step (1) = 81 + 72 + 72 = 225
--------------------------------------------------------------
Step (2) - The number of 3-digit integers with exactly two 7's
Case i: 7 7 _
We can select the 3rd digit in 9 ways (any digit EXCEPT 7)
So, total = 9

Case ii: 7 _ 7
We can select the 2nd digit in 9 ways (any digit EXCEPT 7)
So, total = 9

Case iii: _ 7 7
We can select the 1st digit in 8 ways (any digit EXCEPT 7 and 0)
So, total = 8

So, total for step (2) = 9 + 9 + 8 = 26
--------------------------------------------------------------
Step (3) - The number of 3-digit integers with exactly three 7's
7 7 7
There's only 1 such number
So, total for step (3) = 1
--------------------------------------------------------------

So, the GRAND TOTAL = 225 + 26 + 1 = 252

Phew!! Now I need a nap!

Will there be any question

Will there be any question related to permutation in GMAT exam because I could not see any tutorial for the same in these videos?

True permutation-type

True permutation-type questions are rare on the GMAT. IF you get a permutation question, it can easily be solved using the Fundamental Counting Principle (https://www.gmatprepnow.com/module/gmat-counting/video/775)

Hi Brent,

Hi Brent,

The scope of your proposition (permutation questions can be solved using Fundamental Counting Principle) is restricted to GMAT or extends to the whole subject of permutation. Just asking in case in future I m required to study it as FCP has made my life far easier and I m not willing to move to permutation as such.

Thanks.

Yes, all permutation

Yes, all permutation questions can be solved using Fundamental Counting Principle.

sorry to ask a very basic

sorry to ask a very basic question.. is zero a positive integer? i thought it was neither positive nor negative, but an even integer. and hence formed the 3 digits from 1 to 9 rather than 0 to 9. Googled it but many discrepancies.

I think you may be confusing

I think you may be confusing digits with integers. The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

We use digits to create integers.

So, for example, 507 is a positive integer, which is comprised of three digits: 5, 0 and 7. So, 507 meets the condition of being a positive integer with at least one 7.

Aside: The integer 0 is neither positive or negative.

From intuition I would have

From intuition I would have guessed that there would be 999 possible 3 digit integers, instead of the correct answer of 900... what am I missing?

You are including numbers

You are including numbers where the first one or two digits are zero. For example, 001, 034, etc.

However, on the GMAT, numbers like 001 and 034 are not considered 3-digit numbers, since 001 is really just 1, and 034 is really just 34.

What about how many 3 digit

What about how many 3-digit positive integers with at most two 7's?

So, you're asking how many 3

So, you're asking how many 3-digit numbers have zero, one or two 7's.

So, # of 3-digit positive integers with at most two 7's = (TOTAL # of 3-digit integers) - (# of 3-digit integers with three 7's)

There are 900 3-digit numbers (from 100 to 999 inclusive).

There is 1 3-digit number with three 7's (777)

So, So, # of 3-digit positive integers with at most two 7's = 900 - 1 = 899

What should be the basic

What should be the basic approach to determine a 3 digit positive number with at most one 7.

I'd say the best way is to

I'd say the best way is to recognize the following:

# of 3-digit positive integers with at most one 7 = (TOTAL # of 3-digit integers) - (# of 3-digit integers with three 7's) - (# of 3-digit integers with two 7's)

This quite a bit of work. More work than would be reasonable for a GRE question.

Thanks. I was wondering why

Thanks. I was wondering why none of the practice questions in any of the modules was an "at most" question.

Although your method is easier, I thought of it this way:

1) There are 900 three-digit numbers.

2) In each set of a hundred, there are 10 numbers with 7 as the units digit (7, 17, 27, etc.), so there are 10 x 9 = 90 such numbers.

3)In each set of a hundred, there are 10 numbers with 7 as the tens unit (70-79, 170-179, etc.), so there are 10 x 9 = 90 such numbers. HOWEVER, in each set of a hundred, we've already counted 1 of those numbers already in the units set above (eg. 77), so the total number of three-digit numbers with a 7 as the tens digit is 90 - 9 = 81.

3) Lastly, there are 100 integers with a 7 as the hundreds digit (700-799), BUT we've already counted all the 10 numbers in that set where 7 is in the tens digit (770-779) and the remaining numbers where 7 is in the ones digit (9 remaining numbers since we've already accounted for 777). Thus, the total number of three-digit integers with at least one 7 is 100+81+81 = 252.

I know that's slower than the way you showed, but is that a valid method?

That's a perfectly valid

That's a perfectly valid approach - nice work!

Please elaborate how to solve at most one 7

Are you asking me to elaborate on a former solution I've made, or are you asking me solve a different question altogether?

Are you asking me to solve the question: How many positive 3-digit integers contain AT MOST one 7?

Cheers,
Brent

How many positive 3-digit

How many positive 3-digit integers contain AT MOST one 7

Number of 3-digit integers

Number of 3-digit integers contain AT MOST one 7 = (number of 3-digit integers with ZERO 7's) + (number of 3-digit integers with ONE 7)

Number of 3-digit integers with ZERO 7's
Stage 1: Select hundreds digit in 8 ways (can be 1,2,3,4,5,6,8 or 9)
Stage 2: Select tens digit in 9 ways (can be 0,1,2,3,4,5,6,8 or 9)
Stage 3: Select units digit in 9 ways (can be 0,1,2,3,4,5,6,8 or 9)
Total = (8)(9)(9) = 648
------------------------------------

Number of 3-digit integers with ONE 7
These integers can be in the form 7--, -7- and --7

Integers in the form 7--
Stage 1: Select tens digit in 9 ways (can be 0,1,2,3,4,5,6,8 or 9)
Stage 2: Select units digit in 9 ways (can be 0,1,2,3,4,5,6,8 or 9)
Total = (9)(9) = 81

Integers in the form -7-
Stage 1: Select hundreds digit in 8 ways (can be 1,2,3,4,5,6,8 or 9)
Stage 2: Select units digit in 9 ways (can be 0,1,2,3,4,5,6,8 or 9)
Total = (8)(9) = 72

Integers in the form --7
Stage 1: Select hundreds digit in 8 ways (can be 1,2,3,4,5,6,8 or 9)
Stage 2: Select tens digit in 9 ways (can be 0,1,2,3,4,5,6,8 or 9)
Total = (8)(9) = 72

Number of 3-digit integers with ONE 7 = 81 + 72 + 72 = 225
---------------------------------

Number of 3-digit integers contain AT MOST one 7 = 648 + 225
= 873

Cheers,
Brent

Thank you very much. I did

Thank you very much. I did poor in counting part. I would like to ask if i could well study Probability part to excel all GMAT questions in counting.

Many probability questions

Many probability questions can be solved using either counting strategies or probability rules.

Cheers,
Brent

Hi Brent,

Hi Brent,

Is there any caveat by which you could quickly identify when to use the strategy 1 and when to use strategy 2?

When it comes to questions

When it comes to questions involving "at least," Strategy 2 will typically be the fastest strategy.

If we begin our solution using Strategy 1, we quickly see that we'll need to consider several different cases, and each of these cases will require various calculations. Strategy 2 doesn't suffer from the "multiple cases" issue.

You can also calculate the

You can also calculate the "ignore restriction" by considering that there are 999 numbers with less than 4 digits and subtract 99 of them that have less than 3 digits.
Or take 10^3 combinations of 10 ways that ignores the restriction (3-digit number) and subtract 1x10^2 combinations of 10 ways that break the restriction (3-digit number hence 024,001,099,000,etc).

That's a nice way to

That's a nice way to determine the number of 3-digit integers!

Hey there Brent,

Hey there Brent,

So prior to watching the video I solved the question by doing 7x6x6 = 252. Was that just pure luck or can I apply this method to similar questions?

It's hard to tell what your

It's hard to tell what your method is.
How did you arrive at the three values in your product?
That is, what do 7, 6 and 6 represent?

The 0 got me! Totally forgot

The 0 got me! Totally forgot to remove it from the 1st place! At least next time, I'll remember it for sure!

That's a VERY common error :-

That's a VERY common error :-)

Hey I did it right but

Hey I did it right but through another way:

I used 9 choose 7 and times the result by 7, is this a valid approach?

While it's true that 9C7 x 7

While it's true that 9C7 x 7 = 252 (the correct answer), I can't tell whether your solution is valid.