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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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Updated on: 21 Aug 2019, 12:22
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Difficulty:

65% (hard)

Question Stats:

59% (02:12) correct 41% (02:04) wrong based on 158 sessions

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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

Originally posted by chondro48 on 03 Aug 2019, 02:25.
Last edited by chondro48 on 21 Aug 2019, 12:22, edited 3 times in total.
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How many different 3-digit integers have exactly two d  [#permalink]

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Updated on: 21 Aug 2019, 12:22
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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)

Originally posted by chondro48 on 03 Aug 2019, 02:54.
Last edited by chondro48 on 21 Aug 2019, 12:22, edited 1 time in total.
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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03 Aug 2019, 02:42
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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03 Aug 2019, 02:51
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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

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

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Updated on: 03 Aug 2019, 03:11
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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

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

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Updated on: 09 Sep 2019, 15:32
.

Originally posted by chondro48 on 03 Aug 2019, 03:04.
Last edited by chondro48 on 09 Sep 2019, 15:32, edited 4 times in total.
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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03 Aug 2019, 03:12
I missed the question completely.

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

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03 Aug 2019, 07:42
1
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

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

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07 Aug 2019, 21:16
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

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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08 Aug 2019, 00:31
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]

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09 Sep 2019, 06:46
What about negative 3 digit integers
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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09 Sep 2019, 09:37
1
AAB or ABA - $$9C_1*2C_1*9C_1$$ = 168
BAA - $$9_C_1*9_C_1$$ = 81

Total - 243, Hence E.
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Re: How many different 3-digit integers have exactly two d  [#permalink]

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13 Sep 2019, 01:14
So are there 243+243 = 486 numbers, negative and Positive put together
Re: How many different 3-digit integers have exactly two d   [#permalink] 13 Sep 2019, 01:14