GMAT question - symantic problem?

[Misfire]

This was in a GMAT book, and I think the wording is a little ambiguous. I can see how to get two answers depending on how I read it. Thoughts?

Entries in a particular lottery game are made up of three digits, each 0 through 9. If the order of digits in the entries matters, how many different possible entries exist in which all three digits are not equal?

[ ]516
[ ]720
[ ]989
[ ]990
[ ]1321

[madnak]

I get 720, and it doesn't seem close. The first digit is one of the ten, the second number can be any of the other 9, and the third can be any of the remaining 8. 10*9*8=720. Where is the ambiguity you're seeing?

[Sephus]

the question could mean "how many different possible entries exist in which it's not the case that all three digits are equal?"

like 001. you could say "are all three digits equal? no. therefore, all three digits are not equal."

in this case, you get 990.

[Misfire]

That's what I got too, but the answer explanation says 990. There are 1000 ways to pick numbers, but 10 of them have all three with the same number (1 1 1, 2 2 2, etc).

I would buy this if the question read, "in which not all three digits are equal" rather than "in which all three digits are not equal."

[madnak]

Oh, another way to prove it is to look at how many repeats there are for each digit.

So for 1, there's 011, 111, 211, 311... and there's 110, 111, 112, 113... and there's 101, 111, 121, 131... There are ten numbers in each group. 30 numbers altogether. But 111 appears in all three groups - we don't want to count it three times, only once, so we really have 28 unique numbers.

Since there are only 3 digits in each number, only one of the digits can repeat. Therefore, every digit will repeat the same number of times through the whole 1000 possible 3-digit numbers. We know this number is 28, and there are 10 digits, so the total number of repeats is 280. There are 1000 numbers altogether, so there are 1000-280=720 numbers that don't repeat.

[madnak]

[ QUOTE ]
That's what I got too, but the answer explanation says 990. There are 1000 ways to pick numbers, but 10 of them have all three with the same number (1 1 1, 2 2 2, etc).

I would buy this if the question read, "in which not all three digits are equal" rather than "in which all three digits are not equal."

[/ QUOTE ]

Effing dumb. [censored] question, period. What kind of lame question would that be, even if it were worded properly?

(All three digits are not) (equal) is clumsy language. (All three digits are) (not equal) isn't great either, but is the better interpretation IMO. Just poorly written. Ass.

[Misfire]

Those were my thoughts. Just wanted to make sure I wasn't just misreading.

[Phil153]

"in which there are at least two unique digits"

How effin hard is that? The way the question is written 720 is more correct than 990 imo.

[mickeyg13]

I agree it's poorly worded, but it's incorrect to use the 720 interpretation. The question said nothing about pairs of digits, so you have no reason to disallow pairs of digits.

[Fly]

[ QUOTE ]

If the order of digits in the entries matters, how many different possible entries exist in which all three digits are different?


[/ QUOTE ]

This is what they would have asked if they were looking for 10*9*8.