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How many different 3-digit integers have exactly two d

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How many different 3-digit integers have exactly two d  [#permalink]

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New post Updated on: 03 Aug 2019, 03:56
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How many different 3-digit integers have exactly two different digits?

(A) 1000
(B) 648
(C) 504
(D) 352
(E) 243

Give +1 kudo if you like this question

Originally posted by chondro48 on 03 Aug 2019, 03:25.
Last edited by chondro48 on 03 Aug 2019, 03:56, edited 2 times in total.
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 03 Aug 2019, 03:54
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Prasannathawait wrote:
IMO the answer has to be 162.
9x9x2=162


First digit can be any among the 9 digits
Second cannot be 1 digit that is in first place, hence 9 possible options.
Third has to be one of the two digits of first or second otherwise 3 unique digits will be there.


Hi, you seems to miss the point that exactly two different digits can mean:

AAB, ABA, ABB.

I believe you missed out the third part. Have you carefully taken into account that?

AAB = 9*1*9 =81
ABA = 9*9*1 =81
ABB = 9*9*1 =81

Sum of all is 243 (E)

Hit +1 kudo if you like my solution

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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 03 Aug 2019, 03:42
IMO the answer has to be 162.
9x9x2=162


First digit can be any among the 9 digits
Second cannot be 1 digit that is in first place, hence 9 possible options.
Third has to be one of the two digits of first or second otherwise 3 unique digits will be there.
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 03 Aug 2019, 03:51
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Number of 3-digit integers that have exactly two different digits = Total number of 3 digit integers \(-\) Number of 3 digit integers with 0 different digits (All digits are same) \(-\) Number of 3 digit integers with 3 different digits (All digits are different)

Total number of 3 digit integers = 999-100+1 = 900

Number of 3 digit integers with 0 different digits (All digits are same) = 9*1*1 = 9 (These are 111, 222, 333 ..... 999)

Number of 3 digit integers with 3 different digits (All digits are different) = 9*9*8 = 648 (First digit can be 1-9, second digit can be 0-9 except the one used for first digit, third digit can be 0-9 except the digits used for first and second digits)

So, Number of 3-digit integers that have exactly two different digits = 900-9-648 = 243

Answer is (E)
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How many different 3-digit integers have exactly two d  [#permalink]

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New post Updated on: 03 Aug 2019, 04:11
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We can translate different 3-digit integers have exactly two different digits as this:

All 3 digit integers - all 3 different digit integers - all 3 same digit integers = all 3 digit integers with two different digits, so:

All 3 digit integers = 9 * 10 * 10
all 3 different digit integers = 9 * 9 * 8
all 3 same digit integers = 111, 222, 333... = 9

all 3 digit integers with two different digits = 900 - 648 - 9 = 243

(E) is our answer

Originally posted by Mizar18 on 03 Aug 2019, 04:01.
Last edited by Mizar18 on 03 Aug 2019, 04:11, edited 1 time in total.
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How many different 3-digit integers have exactly two d  [#permalink]

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New post Updated on: 03 Aug 2019, 08:57
Anyone, try this question with timer, leave out good explanation, and kindly leave +1 kudo if this question is quite good.

Originally posted by chondro48 on 03 Aug 2019, 04:04.
Last edited by chondro48 on 03 Aug 2019, 08:57, edited 3 times in total.
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 03 Aug 2019, 04:12
I missed the question completely.
:cry:

chondro48 wrote:
Prasannathawait wrote:
IMO the answer has to be 162.
9x9x2=162


First digit can be any among the 9 digits
Second cannot be 1 digit that is in first place, hence 9 possible options.
Third has to be one of the two digits of first or second otherwise 3 unique digits will be there.


Hi, you seems to miss the point that exactly two different digits can mean:

AAB, ABA, ABB.

I believe you missed out the third part. Have you carefully taken into account that?

AAB = 9*1*9 =81
ABA = 9*9*1 =81
ABB = 9*9*1 =81

Sum of all is 243 (E)

Hit +1 kudo if you like my solution

Posted from my mobile device

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Your suggestions will be appreciated: https://gmatclub.com/forum/your-one-advice-could-help-me-poor-gmat-scores-299072.html

1) Gmat prep: 620 Q48, V27
2) Gmat prep: 610 Q47, V28
3) Gmat prep: 620 Q47, V28
4) Gmat prep: 660 Q47, V34
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 03 Aug 2019, 08:42
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chondro48 wrote:
exc4libur, try this question and leave +1 kudo if this question is quite good.


[method 1]
three-digit numbers with 2 different digits has 3 cases:
XXY: hundreds ≠ 0 so 9 choices • tens = hundreds = 1 choice • units ≠ other digits = 10-1 = 9 choices … 9*1*9=81
XYX: hundreds ≠ 0 so 9 choices • units = hundreds = 1 choice • tens ≠ other digits = 10-1 = 9 choices … 9*1*9=81
YXX: hundreds ≠ 0 so 9 choices • tens ≠ hundreds = 9 choices • units = hundreds = 1 choice … 9*9*1=81

total: 81+81+81=243

[method 2]
3d numbers - 3d triplets - 3d's with different digits = 3d numbers with 2 different digits:
3d numbers = 9*10*10 = 900 or (range between 999-100 inclusive is 999-100+1=900)
3d triplets = 9*1*1 (9 because hundreds ≠ 0) = 9
3d's with different digits = 9*9*8= 81*8 = 648
total: 900-9-648=243

Answer (E).
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 07 Aug 2019, 22:16
chondro48 wrote:
How many different 3-digit integers have exactly two different digits?

(A) 1000
(B) 648
(C) 504
(D) 352
(E) 243

Give +1 kudo if you like this question


Approach: Total - unfavourable cases with zero

Total: 10C2*2C1*3

Cases which start with zero:
1) 1 zero and 2 any other number (first digit can't be zero, everything else is acceptable): 9
2) 2 zero and 1 any other number (other number has to be in hundredth place, everything else is not acceptable): 9*2

Therefore answer: 10C2*2C1*3 - (9 +9*2)
243
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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New post 08 Aug 2019, 01:31
chondro48 wrote:
Prasannathawait wrote:
IMO the answer has to be 162.
9x9x2=162


First digit can be any among the 9 digits
Second cannot be 1 digit that is in first place, hence 9 possible options.
Third has to be one of the two digits of first or second otherwise 3 unique digits will be there.


Hi, you seems to miss the point that exactly two different digits can mean:

AAB, ABA, ABB.

I believe you missed out the third part. Have you carefully taken into account that?

AAB = 9*1*9 =81
ABA = 9*9*1 =81
ABB = 9*9*1 =81

Sum of all is 243 (E)

Hit +1 kudo if you like my solution

Posted from my mobile device


Hi,
I solved the same way, but I took one additional case apart from the three cases (AAB, ABA, ABB) and that is BAA.
For example;
AAB= 223
ABA= 232
ABB= 233
BAA= 322
Can you please explain where did I go wrong?
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Re: How many different 3-digit integers have exactly two d   [#permalink] 08 Aug 2019, 01:31
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