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permutation gmat

by rommysingh » Sun Oct 25, 2015 3:04 pm
Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216
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by Brent@GMATPrepNow » Sun Oct 25, 2015 3:21 pm
rommysingh wrote:Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216
Take the task of creating suitable 3-digit numbers and break it into stages.

Stage 1: Select the single digit that will be different from the other 2 digits
We can choose any of the following 9 digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, we can complete stage 1 in 9 ways

Stage 2: Choose where that single digit (selected above) will be placed
There are 3 places (hundreds position, tens position or units position) where we can place this digit.
So, we can complete stage 2 in 3 ways.

At this point, we have 2 spaces left, and these spaces will be filled by the same digit.

Stage 3: Select the digit to occupy those two remaining spaces
There are 8 remaining digits from which to choose, so we can complete this stage in 8 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create the required 3-digit number) in (9)(3)(8) ways ([spoiler]= 216 ways[/spoiler])

Answer: E
--------------------------

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Cheers,
Brent
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by Matt@VeritasPrep » Sun Oct 25, 2015 10:03 pm
Another approach is to conceptualize the numbers first, then find the values that find your conception.

Since we need two identical digits and one distinct digit, our number should be of the form

AAB
ABA
BAA

Since we can't use 0 for any digit, we have 9 values to choose for A: 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Since B ≠ A, after we choose a value for A, we'll only have 8 values left for B.

So for AAB, we have 9 options for the first A, 1 option for the second A (it has to match the first one!), and 8 options for B, for a total of 9*8 = 72 numbers.

ABA and BAA will also have 72, so we're done: the answer is 72 + 72 + 72, or 216.
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by GMATGuruNY » Mon Oct 26, 2015 2:39 am
rommysingh wrote:Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216
Alternate approach:

Integers with exactly 2 digits the same = Total integers - Integers with all 3 digits the same - Integers with all 3 digits different.

Total integers:
Number of options for the hundreds digit = 9. (Any digit but 0.)
Number of options for the tens digit = 9. (Any digit but 0.)
Number of options for the units digit = 9. (Any digit but 0.)
To combined these options, we multiply:
9*9*9.

Integers with all 3 digits the same:
111, 222, 333, 444, 555, 666, 777, 888, 999.
Number of options = 9.

Integers with all 3 digits different:
Number of options for the hundreds digit = 9. (Any digit but 0.)
Number of options for the tens digit = 8. (Any digit 1-9 other than the digit already used.)
Number of options for the units digit = 7. (Any digit 1-9 other than the two digits already used.)
To combine these options, we multiply:
9*8*7.

Thus:
Integers with exactly 2 digits the same = (9*9*9) - 9 - (9*8*7) = 9(81-1-56) = 9(24) = 216.

The correct answer is E.
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by hoppycat » Wed May 31, 2017 5:12 am
I have a question that isnt about the question here but it's bugging me
Can we have zeros at the front of a 3 digit number? for example does 043 count?
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by Brent@GMATPrepNow » Wed May 31, 2017 5:21 am
hoppycat wrote:I have a question that isnt about the question here but it's bugging me
Can we have zeros at the front of a 3 digit number? for example does 043 count?
No, 043 doesn't count as a 3-digit number.
The 3-digit integers range from 100 to 999 inclusive.
Likewise, the 4-digit integers range from 1000 to 9999 inclusive.

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by Zoser » Mon Dec 04, 2017 5:57 am
Take the task of creating suitable 3-digit numbers and break it into stages.

Stage 1: Select the single digit that will be different from the other 2 digits
We can choose any of the following 9 digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, we can complete stage 1 in 9 ways

Stage 2: Choose where that single digit (selected above) will be placed
There are 3 places (hundreds position, tens position or units position) where we can place this digit.
So, we can complete stage 2 in 3 ways.

At this point, we have 2 spaces left, and these spaces will be filled by the same digit.

Stage 3: Select the digit to occupy those two remaining spaces
There are 8 remaining digits from which to choose, so we can complete this stage in 8 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create the required 3-digit number) in (9)(3)(8) ways
Hi Brent,

Why can't I just do the following?

Choose the 1st number: we can do this in 9 ways.
Choose the 2nd number: 8 ways
Choose the 3rd number: 1 yay

Total= 9*8= 72

What is wrong in this approach?
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by Brent@GMATPrepNow » Mon Dec 04, 2017 7:38 am
Zoser wrote:
Take the task of creating suitable 3-digit numbers and break it into stages.

Stage 1: Select the single digit that will be different from the other 2 digits
We can choose any of the following 9 digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, we can complete stage 1 in 9 ways

Stage 2: Choose where that single digit (selected above) will be placed
There are 3 places (hundreds position, tens position or units position) where we can place this digit.
So, we can complete stage 2 in 3 ways.

At this point, we have 2 spaces left, and these spaces will be filled by the same digit.

Stage 3: Select the digit to occupy those two remaining spaces
There are 8 remaining digits from which to choose, so we can complete this stage in 8 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create the required 3-digit number) in (9)(3)(8) ways
Hi Brent,

Why can't I just do the following?

Choose the 1st number: we can do this in 9 ways.
Choose the 2nd number: 8 ways
Choose the 3rd number: 1 yay

Total= 9*8= 72

What is wrong in this approach?
First, in your second step/stage, you say that you can complete that stage in 8 ways.
This assumes that the 2nd digit is different from the 1st digit.
However, the first 2 digits can be equal (as in, 117 and 884)

Next, in your third step/stage, you say that you can complete that stage in 1 way.
How did you get 1?

Cheers.
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by Zoser » Mon Dec 04, 2017 7:58 am
Brent@GMATPrepNow wrote:
Zoser wrote:
Take the task of creating suitable 3-digit numbers and break it into stages.

Stage 1: Select the single digit that will be different from the other 2 digits
We can choose any of the following 9 digits: 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, we can complete stage 1 in 9 ways

Stage 2: Choose where that single digit (selected above) will be placed
There are 3 places (hundreds position, tens position or units position) where we can place this digit.
So, we can complete stage 2 in 3 ways.

At this point, we have 2 spaces left, and these spaces will be filled by the same digit.

Stage 3: Select the digit to occupy those two remaining spaces
There are 8 remaining digits from which to choose, so we can complete this stage in 8 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create the required 3-digit number) in (9)(3)(8) ways
Hi Brent,

Why can't I just do the following?

Choose the 1st number: we can do this in 9 ways.
Choose the 2nd number: 8 ways
Choose the 3rd number: 1 yay

Total= 9*8= 72

What is wrong in this approach?
First, in your second step/stage, you say that you can complete that stage in 8 ways.
This assumes that the 2nd digit is different from the 1st digit.
However, the first 2 digits can be equal (as in, 117 and 884)

Next, in your third step/stage, you say that you can complete that stage in 1 way.
How did you get 1?

Cheers.
Brent
I got 1 beacuase after we vhoose the 2nd number, the last number is the same so there is only 1 way for that!
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by Brent@GMATPrepNow » Mon Dec 04, 2017 1:48 pm
Zoser wrote:
I got 1 beacuase after we vhoose the 2nd number, the last number is the same so there is only 1 way for that!
I see.
You have answered a slightly different question.
You have answered the question "How many 3 digit numbers are there such that the last 2 digits are identical AND different from the first digit"
So, your solution only counts numbers like 122, 344, 622, 955, 822, 844 etc.

However, we must also count numbers like 557, 575, 441, 363, 991, etc

Cheers,
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by Scott@TargetTestPrep » Fri Jan 12, 2018 6:32 am
rommysingh wrote:Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216
We have three cases: 1) the hundreds and tens digits are the same and the ones digit is different (XXY), 2) the hundreds and ones digits are the same and the tens digit is different (XYX), and 3) the tens and ones digits are the same and the hundreds digit is different (YXX).

Since no digits are equal to 0, any of the three cases above have an equal number of integers. So if we find the number of integers for one case, we can multiply that by 3 to get the total number of integers where two digits are the same and the third is not. Let's analyze case 1:

There are 9 choices (1 to 9) for hundreds digit, but there is only 1 choice for the tens digit since it has to be the same as the hundreds digit, and there are 8 choices for the ones digit since it has to be a different digit than that of the hundreds and tens digits. Thus the number of integers in case 1 is:

9 x 1 x 8 = 72

Multiply 72 by 3, we will have 216 integers where two digits are the same and the third is not.

Answer: E

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