The Answer Key appears at the end of the review sheet, which is posted on the “Exam Reviews” page.  Let me know if you find any errors, or have any questions.

Best of luck,
Prof. Reitz

1. ibrahimadam193

I have a question about problem 14a.

The problem asks us to find the amount of 4 element lists where the first two elements must be vowels.

The answer I got was $3\cdot 2\cdot 8 \cdot 7$.

My reasoning is that there are three vowels in the set. Two of these vowels must be in the first two positions of this list. In the last two positions can go any of the leftover 7 consonants or the vowel that was not placed in the front of the list. In total, this gives us 8 elements to order in the last two positions. The answer in the answer key suggests that there are only 7 elements to order in the last two positions, which would be true if only consonants could be placed in the last two positions.

Is my thinking wrong?

2. Jonas Reitz

You are absolutely right – the correct answer is $3\cdot 2\cdot 8\cdot 7$, or 336. I’ve updated the Answer Key accordingly.
Nice work!
Prof. Reitz

3. Benjamin Zeng

For problem 11b, the intersect one, shouldn’t the answer be [0,1]. The first one goes from [0,1], the next [0,2] and so on thus they should all include the interval [0,1] unless you’re counting zero as a positive number.

4. Jonas Reitz

Hi Benjamin,
For 11b, note that the positive real numbers includes numbers between 0 and 1 — so, for example, the interval [0, .5] is one of our sets, as is [0, .1] and so on. This means that the only number that is in all of them is 0 itself.
-Prof. Reitz

• Benjamin Zeng

For some reason i was thinking integers. Thank you.