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How many different three-digit numbers can be formed which contain two

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How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 29 Mar 2018, 00:13
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Difficulty:

  55% (hard)

Question Stats:

65% (01:44) correct 35% (02:31) wrong based on 86 sessions

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[GMAT math practice question]

How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?

A. 196
B. 216
C. 243
D. 256
E. 316

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Re: How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 29 Mar 2018, 01:00
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Not sure, but should be: 9*9*1*3!/2!=243
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Re: How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 29 Mar 2018, 02:04
The answer should be (B). 9*8*1*3!/2! = 216

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Re: How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 31 Mar 2018, 05:01
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MathRevolution wrote:
[GMAT math practice question]

How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?

A. 196
B. 216
C. 243
D. 256
E. 316


Start from total of ways to obtain 3 digit numbers : 9 * 10 * 10 = 900

The number of ways we are looking for is :Exactly 2 identical digits and 1 different digit (respecting scope constraint : 0 cannot be first digit)

So let's subtract from 900, cases where all digits are different, and cases where all 3 digits are the same, leaving logically the number of ways we are looking for : 900 - (9 * 9 * 8) - (9 * 1 * 1) = 243

Hope it helps
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Re: How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 31 Mar 2018, 08:34
2
We can see that there can be 3 scenarios:

1. XXY
2. XYX
3. YXX

XX can be given a value between 1-9 in scenarios 1 and 2. and Y can be given any value 0-9 that is not X so we're looking at 9*9=81

Each scenario there can only be 9*9 outcomes. So 3*81=243. C
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Re: How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 01 Apr 2018, 17:46
=>

These three-digit numbers can have one of the forms XXY, XYX and YXX.
Note that 0 cannot be the hundreds digit.

Case 1): XXY
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 2): XYX
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 3): YXX
There are 9 possibilities for Y (Y is not 0), and 9 possibilities for X (X≠Y).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Thus, the total number of possible three-digit numbers is 81 + 81 + 81 = 243.

Therefore, C is the answer.
Answer: C
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Re: How many different three-digit numbers can be formed which contain two  [#permalink]

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New post 19 Dec 2018, 09:44
Hi Terrifficm

suppose the two same digits are x and one different digit is y

Answer is not 216 because, for two same digits we can select from nine digits - 1,2,3,4,5,6,7,8,9. Therefore, it would be 9C1 i.e. 9

and for one different digit , we can select from - 0,1,2,3,4,5,6,7,8,9 i.e 10-1 digit i,e again 9 digits. therefore, it would be 9C1 i.e.9

Again these three digits can be arranged among themselves in three ways for e.g. xxy, xyx, yxx

thus the calculation is 9*9*3 which is 243

trust this helps
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Re: How many different three-digit numbers can be formed which contain two &nbs [#permalink] 19 Dec 2018, 09:44
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