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Answer to the Lecture Question from Last Time
Total number of ways to choose 7 questions:
10
C
7
Total number of ways to choose 7 questions by choosing less
than 4 of the first 6:
6
C
3
.
Now, the second answer does not seem obvious. It should not
be. If you are answering 7 of 10 questions, then you can leave
at MOST 3 of them blank. Thus, if MUST choose at LEAST 3
questions of the first 6, no matter what. So, the only way to
choose 7 out of 10 questions, WITHOUT choosing at least 4 of
these first 6 questions is to choose exactly 3 of them.
The answer to our question is simply the difference of these
two values:
10
C
7

6
C
3
= 100.
Another way to view this question is the following:
We can choose our questions in the following 3 ways:
1) 4 of the first 6, followed by 3 of the last 4
2) 5 of the first 6, followed by 2 of the last 4
3) 6 of the first 6, followed by 1 of the last 4.
We can do #1 in (
6
C
4
) (
4
C
3
) = 60
We can do #2 in (
6
C
5
) (
4
C
2
) = 36
We can do #3 in (
6
C
6
) (
4
C
1
) = 4
We used the multiplication principle for each of these parts
and will use the sum rule to deduce that there are a total of 100
possible test question choices.
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View Full DocumentA common mistake made on this problem is the following:
1)
Choose 4 of the first 6 questions.
2) Then, you have 3 choices to make of the remaining 6
questions.
It seems as if we can do this in (
6
C
4
) (
6
C
3
) = 300 ways.
But, as we can see, this answer is incorrect, by quite a bit.
We are actually counting some of the combinations of
questions more than once, that is why this answer is incorrect.
Consider the combination 1, 2, 3, 4, 5, 7, and 8. We can arrive
at this combination in a couple ways:
1) Pick 1,2,3,4 of the first 6, then 5,7,8 of the remaining Qs.
2) Pick 1,2,3,5 of the first 6, then 4,7,8 of the remaining Qs.
We count these two possibilities separately in the count above.
But, we shouldn’t. Thus, we are over counting and end up with
an incorrect answer.
Combinations with Repetition
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 Spring '09

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