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Of the three-digit positive integers that have no digits

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Of the three-digit positive integers that have no digits  [#permalink]

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New post 31 Jan 2007, 06:12
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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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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 27 Jan 2012, 00:25
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Baten80 wrote:
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.


I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with.

As for the above problem. One can also use combination approach.

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 a three digit integer: XXY.
Choosing the digit for X - 9 ways;
Choosing the digit for Y - 8 ways;
# of permutations of 3 digits in XXY - 3!/2! (as 2 X's are identical);

Total: 9*8*3!/2!=216.

Answer: E.

Hope it helps.
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Re: PS - Digits (Very Tough one)  [#permalink]

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New post 31 Jan 2007, 07:21
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TOUGH GUY 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?


Assume: a,b,c is the digit and a,b,c not= 0

Thus a,b,c could be 1, 2, 3, 4, 5, 6, 7, 8, and 9

Count the possible way to get three digit number abc.

a = b, and c must be different from a, b
Thus, there are 3 possible ways of digit arrangement: aac, aca, caa

Case I: aac
=> (digit 1st) x (digit 2nd) x (digit 3rd)
=> 9 x 1 x 8 {pick any number from group = 9 possible ways} x {pick number the same as the first pick = 1 way} x {pick any number from the rest = 8 possible ways}

= 9 x 1 x 8 = 72 possible ways

Case II: aca
=> same as case I you have 72 possible ways

Case III: caa
=> same as case I you have 72 possible ways

total of this set of number = 72 + 72 + 72 = 216

E) is the answer

(I got this question the first time 72 but I found that is question needs a little more work to do :wink: )
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Re: PS - Digits (Very Tough one)  [#permalink]

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New post 31 Jan 2007, 22:49
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TOUGH GUY 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


three-digit positive integer where two digits that are equal to each other and the remaining digit different from the other two
A B C -three digits
possible combinations
AAC
ACA
CAA
1-1-(2,3,4,5,6,7,8,9) total 8 combinations
1-((2,3,4,5,6,7,8,9)-1 total 8 combinations
(2,3,4,5,6,7,8,9)-1,1 total 8 combinations
3*8=24
and so on

we have 9 digits with 24 possible comb for each
total 9*24=216
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  [#permalink]

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New post 01 Feb 2007, 01:13
Zero can't be a digit so the not equal can be selected 9 ways and equal can be selected 8 ways or vice versa.

If Number is XXY then there are 8*9=72 possibilities
If Number is XYX then there are 8*9 =72 possibilities
If Number is YXX then there are 8*9 =72 possibilities
Total 216
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 27 Jan 2012, 00:06
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 27 Jan 2012, 05:13
good question...

good explanation too..
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 10 Feb 2014, 06:02
Bunuel wrote:
Baten80 wrote:
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.


I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with.

As for the above problem. One can also use combination approach.

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 a three digit integer: XXY.
Choosing the digit for X - 9 ways;
Choosing the digit for Y - 8 ways;
# of permutations of 3 digits in XXY - 3!/2! (as 2 X's are identical);

Total: 9*8*3!/2!=216.

Answer: E.

Hope it helps.


Is it possible to solve the inverse? Say TOTAL - (Cases were ALL are equal ) - (Cases were ALL are different)?

I'm trying to take a stab at it but I'm not able to come with the answer

Is it 9^3 - (9*3) - (9*8*7) ?

Thanks
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 10 Feb 2014, 06:11
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jlgdr wrote:
Bunuel wrote:
Baten80 wrote:
many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems.


I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with.

As for the above problem. One can also use combination approach.

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 a three digit integer: XXY.
Choosing the digit for X - 9 ways;
Choosing the digit for Y - 8 ways;
# of permutations of 3 digits in XXY - 3!/2! (as 2 X's are identical);

Total: 9*8*3!/2!=216.

Answer: E.

Hope it helps.


Is it possible to solve the inverse? Say TOTAL - (Cases were ALL are equal ) - (Cases were ALL are different)?

I'm trying to take a stab at it but I'm not able to come with the answer

Is it 9^3 - (9*3) - (9*8*7) ?

Thanks
Cheers
J


Almost nailed it! The # of three-digit integers that have all digits equal is just 9, not 9*3:
111,
222,
...
999.

So, {Total} - {all equal} - {all distinct} = (9*9*9) - 9 - (9*8*7) = 216.

Hope it's clear.
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 24 Sep 2016, 05:08
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Digits can be in the form ABB, BBA, BAB
For ABB - 9 options for A(1-9) and 8 options for B(2-9), since taking B as 1 will give 111 which is not required by question
hence no. of combinations possible 9*8 = 72
similarly for BBA- 9*8=72
and BAB - 9*8 =72
total = 72+72+72 = 216
E is the answer
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 16 Oct 2016, 08:57
We want a number like : XXY, X and Y being digits from 1 to 9.
To determine the number of possibilities, we can use the fundamental counting principle.

1/ Number of possibilities of X : 9
2/ Number of possibilities of Y : 8
3/ Number of ways to arranges XXY : 3!/2! = 3

So, 9*8*3 = 216
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 16 Oct 2016, 09:28
TOUGH GUY 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


Let, here are three places that we have to fill by digits and two of them must be filled by same digit

_ _ _

There are three cases

XXY ===> Total ways = 9*8 = 72 Numbers (X can be chosen 9 ways 1-9, and Y in another 8 ways)
XYX ===> Total ways = 9*8 = 72 Numbers (X can be chosen 9 ways 1-9, and Y in another 8 ways)
YXX ===> Total ways = 9*8 = 72 Numbers (Y can be chosen 9 ways 1-9, and Y in another 8 ways)

Total Numbers = 3*72 = 216

Answer: Option E
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 15 Dec 2017, 14:15
Hi!

Can someone explain why the answer can't work using combination? For this example, I'm looking for the number of variations of AAB. I did 9 pick 2 (since you're using two digits and there's a total of 9 digits you can pick from). Then I multiplied 9 pick 2 (36) by 3!/2! which equals 108.
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 16 Dec 2017, 00:38
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GMATKAY22 wrote:
Hi!

Can someone explain why the answer can't work using combination? For this example, I'm looking for the number of variations of AAB. I did 9 pick 2 (since you're using two digits and there's a total of 9 digits you can pick from). Then I multiplied 9 pick 2 (36) by 3!/2! which equals 108.


9C2 = 36 is the number of pairs of different digits from 9 digits: (1, 2); (1, 3); (1, 4); ...; (8, 9).

Now, one pair, say (1, 2), can give you TWO types of numbers: with two 1 and one 2 and two 2's and one 1: 112 (which in turn gives 112, 121, 211) and 221 (which in turn gives 221, 212, 122). So, you should multiply 9C2*3!/2! further by 2: 9C2*3!/2!*2 = 216.

Hope it's clear.
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 06 Dec 2018, 17:29
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TOUGH GUY 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 (= 216 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 11 Jan 2019, 13:00
Here 2 places can be selected in 3c2 ways and we have 9 options for first digit and 1 for the other similar digit
3rd different digit can be selected in 8 ways
so 3c2*9*1*8=216
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Re: Of the three-digit positive integers that have no digits  [#permalink]

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New post 11 Jan 2019, 13:32
TOUGH GUY 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


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.


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Re: Of the three-digit positive integers that have no digits   [#permalink] 11 Jan 2019, 13:32
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