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Combinations - how many 3 digit numbers?

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Combinations - how many 3 digit numbers? [#permalink] New post 03 Mar 2009, 16:33
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4. How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

a) 648.
b) 504.
c) 576.
d) 810.
e) 672.


------------------------------------------------------------------------
What is the answer and explanation please?
- I have 9 x 8 x 7 = 504, but then also think it can be
9 x 9 x 8 = 648?
Confused ?
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Re: Combinations - how many 3 digit numbers? [#permalink] New post 03 Mar 2009, 17:24
sHould be 9x9x8 = 648.
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Re: Combinations - how many 3 digit numbers? [#permalink] New post 03 Mar 2009, 18:54
Forrester300 wrote:
4. How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

a) 648.
b) 504.
c) 576.
d) 810.
e) 672.


------------------------------------------------------------------------
What is the answer and explanation please?
- I have 9 x 8 x 7 = 504, but then also think it can be
9 x 9 x 8 = 648?
Confused ?
JF


This should be correct: 9 x 9 x 8 = 648.

The 3 digit integers are in xyz form: xyz

1: In x's place 1, 2, 3, 4, 5, 6, 7, 8, and 9 can be placed: so 9 ways
2: In y's place 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, but except x, can be placed: 9 ways
3: In z's place 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, but except x and y, can be placed: 8 ways

So, it is; 9 x 9 x 8 = 648.
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Re: Combinations - how many 3 digit numbers? [#permalink] New post 06 Mar 2009, 07:58
How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

a) 648.
b) 504.
c) 576.
d) 810.
e) 672.

If xyz is the number then, doesnt " the other digits " mean y & z?
IMO it should be 9 * 10 * 9 = 810.

Whats the OA??
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Re: Combinations - how many 3 digit numbers? [#permalink] New post 28 Mar 2010, 04:00
aalriy wrote:
How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

a) 648.
b) 504.
c) 576.
d) 810.
e) 672.

If xyz is the number then, doesnt " the other digits " mean y & z?
IMO it should be 9 * 10 * 9 = 810.

Whats the OA??


This is exactly what I thought! "Other digits" is creating ambiguity here. I thought that the last two digits should be different from each other. So I got 810
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Re: Combinations - how many 3 digit numbers?   [#permalink] 28 Mar 2010, 04:00
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