Of the three-digit positive integers that have no digits

Question

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

Bunuel · Accepted Answer

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

Consider a three-digit integer in the form XXY.

Choices for digit X: 9 (since it can be any digit from 1 to 9). 
  Choices for digit Y: 8 (any digit from 1 to 9, excluding X). 
  Permutations of the digits in XXY: 3!/2! (since the two X's are identical).

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

Answer: E.

Hope this helps.

devilmirror · Answer

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:

)

mohan514 · Answer

good question...

good explanation too..

jlgdr · Answer

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

Bunuel · Answer

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.

GMATinsight · Answer

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

GMATKAY22 · Answer

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.

Bunuel · Answer

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.

BrentGMATPrepNow · Answer

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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GMATGuruNY · Answer

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.

E

BrentGMATPrepNow · Answer

Here's another approach.

There are three possible cases that satisfy the given information. 
case i: The integer is in the form XXY
case ii: The integer is in the form XYX
case iii: The integer is in the form YXX

case i: The integer is in the form XXY 
There are 9 ways to select the first digit (1,2,3,4,5,6,7,8 or 9)
There is 1 way to select the second digit (since the second digit must match the first digit)
There are 8 ways to select the third digit (it can be any digit except the digit we chose for first position)
So the total number of integers in the form XXY = (9)(1)(8) =  72

case ii: The integer is in the form XYX 
There are 9 ways to select the first digit (1,2,3,4,5,6,7,8 or 9)
There are 8 ways to select the second digit (it can be any digit except the digit we chose for first position)
There is 1 way to select the third digit (since the third digit must match the first digit)
So, the total number of integers in the form XYX = (9)(8)(1) =  72

As you might guess, each of the three cases can be accomplished in  72  ways

Answer: (3)( 72 ) = 216

Answer: E

Cheers,
Brent

Uddipan1 · Answer

The best way of solving this is:
1. No of ways of choosing 2 digits, say A and B, from 9 available digits = 9C2 = 36
2. No of ways of choosing the repeating digit from the above A and B = 2
3. No of ways of arranging 3 digits such that 2 are repeating = 3!/2! = 3

Answer = 36 * 2 * 3 = 216

kanwar08 · Answer

Bunuel   KarishmaB

I have one doubt about this question. Fundamentally we are trying to find the number of possible 3-digit integers which have 2 same and one different integer.

I tried using the filling-the-spaces approach i.e a,b,b I am filling these three spaces where first can be done in 9 ways, the second 8 ways, and the last in 1 way. The total comes to be 9*8*1 = 72

Later, since aab can be rearranged in 3!/2! ways, so we multiply with that number.

I am getting confused about whether filling the spaces (w/o considering the order) is equivalent to Permutation or Combination.

Do I always need to consider the arrangement whenever I am using the Filling the Spaces Approach?

Please Guide

KarishmaB · Answer

It is pointless to worry about the terminology, but this is a case of permutations because a different arrangement of the same digits gives us a different number. 677 is different from 767 and different from 776. When you fill in abb, you are counting only 677. Hence you multiply by 3 to count the other two arrangements too. Fundamental counting principle as used by you to fill up the spaces is used to fill up distinct spots. You have used FCP on abb giving you all possible values for abb. But what about bab and bba? It is better to think in terms of what all are you accounting for when you are using a particular method. If I can fill the first spot in 9 ways, all 3 digit numbers starting with 1 to 9 are being account for. If I can fill the second spot in 8 ways, I cannot repeat the first digit. So 67_, 76_, 89_ etc are being account for If I can fill the third spot in only 1 way (same as second spot digit), the I am accounting for 677, 766, 899 etc. But have I accounted for 767? No. What about 776? No. Hence I multiply by 3 to get 216. I have discussed some more such questions here: https://youtu.be/_C-kTA44OxE Answer (E)

shashank07! · Answer

Bunuel  what would have been the answer if this constraint :   no digits equal to zero  was not given

Bunuel · Answer

Consider XXY.

Choices for digit X: 10 (since it can be any digit from 0 to 9). 
 Choices for digit Y: 9 (any digit from 0 to 9, excluding X). 
 Permutations of the digits in XXY: 3!/2! (since the two X's are identical).

Total combinations: 10*9*3!/2! = 270.

However, a three digit number cannot start with 0, so we should subtract: 00? and 0?0 (9 + 9 = 18 cases), as well as 0?? (9 cases).

Total = 270 - 18 - 9 = 243.

Alternatively, you can add to the 216 we got above by including ?00 (9 cases) as well as ?0? and ??0 (9 + 9 = 18 cases) to get 216 + 9 + 18 = 243.

Hope it's clear.

Arjun1001 · Answer

CAN YOU PLEASE EXPLAIN HOW THIS IS DIFFERENT FROM THIS QUESTION. WHY SAME METHOD CANT BE USED FOR THIS ONE. https://gmatclub.com/forum/of-the-three ... 35188.html

Bunuel · Answer

You  can  use the same method.

Total number of three-digit numbers with no zeros = 9 * 9 * 9 = 729 (Each digit from 1–9)

Numbers with all distinct digits with no zeros = 9 * 8 * 7 = 504 (First digit: 1–9, second: 8 remaining, third: 7 remaining)

Numbers with all three digits the same = 9 (111, 222, ..., 999)

So, numbers with exactly two digits the same = 729 – (504 + 9) = 216

P.S. Please avoid using all caps in your posts.

GmatKnightTutor · Answer

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

Doing this indirectly may be helpful.

Total possibilities = 9 x 9 x 9 =  729

minus

All 3 digits are the same =  9  (111, 222,... 999)
All 3 digits are different = 9 x 8 x 7 =  504

=

Numbers in which only 2 digits are the same: 216

Of the three-digit positive integers that have no digits

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Of the three-digit positive integers that have no digits

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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