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How many different three-digit integers have exactly three different

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How many different three-digit integers have exactly three different  [#permalink]

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New post 27 Jan 2019, 18:28
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How many different three-digit integers have exactly three different digits?

(A) 504
(B) 648
(C) 720
(D) 891
(E) 1,000

I'm wondering if the way I did it makes sense...
1) Ignoring the restriction that 0 can't be in the first position: 10P3 = 10*9*8 = 720
2) Remove the possibilities with 0 in the first position (1/10th the total of 10P3): 720/10 = 72
3) Subtract the possibilities with 0 in first position from total: 720 - 72 = 648

Explanation:
There are nine different possibilities for the first digit: anything between 1 and 9. The tricky part of this question is moving to the second digit. Again, there are nine possibilities for the second digit: any number between 0 and 9, not including the one using for the first digit. Remember, a three-digit number can have zero as a digit, just not as the first digit. Finally, the third digit has eight possibilities: anything between 0 and 9 except for the two digits used already. The number of possible integers, then, is: 9 × 9 × 8 = 648, choice (B).
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Re: How many different three-digit integers have exactly three different  [#permalink]

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New post 27 Jan 2019, 18:35
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The possibility of being a three digit integer is when we don't consider the first digit to be 0
For the second we can consider 0

Thus
9*9*8
648
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How many different three-digit integers have exactly three different  [#permalink]

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New post 03 Aug 2019, 01:25
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How many different three-digit integers have exactly three different digits?

(A) 504
(B) 648
(C) 720
(D) 891
(E) 1,000

Source: Math Bible (Jeff Sacksmann)
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How many different three-digit integers have exactly three different  [#permalink]

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New post 03 Aug 2019, 03:05
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How many different three-digit integers have exactly three different digits?

Let the number =abc; Notice that a cant take digit 0 as it’s a three-digit number. Digit can’t be reused

Now a has 9 possibilities , b has 9 possibilities as we can’t reuse any digit, lastly c has 8 possibilities

.: Number of different three digit integer = 9•9•8 = 81•8 = 648

Smack that B! :)

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How many different three-digit integers have exactly three different  [#permalink]

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New post Updated on: 17 Aug 2019, 08:23
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Rohanborude,

Let the number be XYZ;

X has 9 possibilities (0 is not allowed).
Y has 9 possibilities (number in X cannot be repeated).
Z has 8 possibilities (numbers in both X and Y cannot be repeated).

Number of different three digit integer = 9*9*8 = 648. Answer is (B)

Smack that +1 kudo below if you like my explanation

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Originally posted by chondro48 on 03 Aug 2019, 03:19.
Last edited by chondro48 on 17 Aug 2019, 08:23, edited 1 time in total.
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Re: How many different three-digit integers have exactly three different  [#permalink]

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New post 03 Aug 2019, 07:24
DarkHorse2019 wrote:
How many different three-digit integers have exactly three different digits?

(A) 504
(B) 648
(C) 720
(D) 891
(E) 1,000

Source: Math Bible (Jeff Sacksmann)


Asked: How many different three-digit integers have exactly three different digits?

For hundred digit, number of choices = 9
For tenth digit, number of choices = 9 (can not use hundred digit again)
For units digit, number of choices =8. (can not use hundred and tenth digits again)

Number of different 3 digit integers = 9*9*8 = 648


IMO B
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Re: How many different three-digit integers have exactly three different  [#permalink]

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New post 03 Aug 2019, 08:02
DarkHorse2019 wrote:
How many different three-digit integers have exactly three different digits?

(A) 504
(B) 648
(C) 720
(D) 891
(E) 1,000

Source: Math Bible (Jeff Sacksmann)


given: there are total of 10 digits (0,1,2,3,4,5,6,7,8,9)

thousands: 9 choices (because you can't have 0)
hundreds: 9 choices (10-1=9 digits left, since we used a digit for the thousands)
units: 8 choices (10-2=8 digits left, since we used a digit for the thousands and another for the hundreds)
total three-digit integers with different digits: 9*9*8=81*8=648

Answer (B).
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Re: How many different three-digit integers have exactly three different  [#permalink]

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New post 03 Aug 2019, 11:13
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Re: How many different three-digit integers have exactly three different  [#permalink]

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New post 03 Aug 2019, 12:08
first digit = 9
second digit = 9 ' 0' also possible
third digit ; 8
9*9*8 ; 648
IMO B

energetics wrote:
How many different three-digit integers have exactly three different digits?

(A) 504
(B) 648
(C) 720
(D) 891
(E) 1,000

I'm wondering if the way I did it makes sense...
1) Ignoring the restriction that 0 can't be in the first position: 10P3 = 10*9*8 = 720
2) Remove the possibilities with 0 in the first position (1/10th the total of 10P3): 720/10 = 72
3) Subtract the possibilities with 0 in first position from total: 720 - 72 = 648

Explanation:
There are nine different possibilities for the first digit: anything between 1 and 9. The tricky part of this question is moving to the second digit. Again, there are nine possibilities for the second digit: any number between 0 and 9, not including the one using for the first digit. Remember, a three-digit number can have zero as a digit, just not as the first digit. Finally, the third digit has eight possibilities: anything between 0 and 9 except for the two digits used already. The number of possible integers, then, is: 9 × 9 × 8 = 648, choice (B).

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Re: How many different three-digit integers have exactly three different   [#permalink] 03 Aug 2019, 12:08
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