Another reading of this question would give 6720. For many teams, the positions are important. There are 8 choices for position a, 7 choices for position b, etc, giving 6720 possibilities. The issue here and in the other questions isn't really which answer is "best" --- I'd have agreed that 56 was the "best" answer, that is, the one they wanted. The problem is that these questions are off---they're constructed and phrased poorly enough to add mud to people's thinking. Muddy textbooks, muddy teaching, muddy tests---unfortunately, that's all standard fare. It's bad for the good students and the struggling students alike. Remember, this is a test that was supposed to have been passed by all NY high school students to graduate, not just people with a special interest in math. Bill Ps. After my previous post, I looked again and realized that in the stem-leaf question, the "key" was not the answer key but the clue to the meaning: it says key 4|3 = 43. That made it more obvious as just a list of numbers sorted into bins according to the first digit. However, the first digits that occurred in the diagram are 4,6,7,9, which seems weird: that's not how you would construct a bin diagram like this. On Friday, July 4, 2003, at 01:05 PM, Henry Baker wrote:

At 12:54 PM 7/4/03 -0400, Bernie Cosell wrote:

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On 4 Jul 2003 at 9:13, Henry Baker wrote:

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At 11:11 AM 7/4/03 -0400, William Thurston wrote:

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20. How many different five-member teams can be made from a group of eight students, if each student has an equal chance of being chosen? (1) 40 (2) 56 (3) 336 (4) 6720

I think the correct answer is not among these choices: 1 It's very poorly phrased, being neither colloquial English or typical mathematical/combinatorial English.

I agree that there is a reading that would give the choice "1", but since that choice is not available, you have to assume that that interpretation is not intended.

Could you elaborate on how to get the choice '1'? Hmm...well... I guess you could say that you can only make *ONE* five-man team, but you could do it in 56 different ways... Is that what you had in minde?

Yes. Once you had made one team, you don't have enough people left over to make another.

BTW, the "equal chance" bit is a red herring. The probabilities don't matter, so long as none of them are zero. I think that this is thrown in there to indicate that you should put on your "combinatorial & probability" hat for this question.

Yes. I'm curious whether the "equal probability" red herring was thrown in to deliberately mislead, or because the exam makers thought it was a clue as to what hat to use, or because they thought it's a necessary qualifier for these kinds of problems. In any case, I don't get a high impression of the exam constructors' clarity of thought.