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Math Challenge I-B: Counting and Probability

 
 
LinAmber的头像
Math Challenge I-B: Counting and Probability
LinAmber - 2019年04月30日 Tuesday 16:07
 

I am confused about these questions:
3.27, 
3.28,
In 4.1: 4.4(b),
In 4.3: 4.25,
In 4.3: 4.30.

 
ReynosoDavid的头像
Re: Math Challenge I-B: Counting and Probability
ReynosoDavid - 2019年05月1日 Wednesday 11:04
 

Hi Amber. Here are some hints for the problems you posted. Remember you can look at the full solutions of all homework problems once you submit the quiz for that section. Letting us what you have tried can help us give you better hints as well.


3.27: The words do not need to be real English words. The only thing you want to make sure is to not have two of the same letter in a row. So $ABABA$ is a valid word, while $AABAB$ is not because it has two $A$'s in a row. To count them do one letter at a time: How many choices do you have for the first letter? For the second? For the third?

3.28: You want to figure out in how many ways you can pick $6$ out of $51$ possible numbers, where the order does not matter. Is this a permutation or a combination?

4.4(b) follows the same approach as 4.4(a). This time, however, the first student picks first (which can be done in $\binom{6}{3}$ ways), then gives the books to the second student (who now has $8 + 3 = 11$ books), and then the second student picks three books among all the books he has (which can be done in $\binom{11}{3}$ ways). So this book exchange can happen in $\binom{6}{3} \times \binom{11}{3}$ different ways.

4.25: Pretend the Klingons, Vulcans and Andorians are making the teams themselves. The Klingons have to be in different groups, so they may as well be the ones to choose. In how many ways can the first Klingon choose one Vulcan and one Andorian? What about the second Klingon? The third?

4.30: is similar to the other "line people in a row" problems. In previous problems when we want a couple to be together, we think of them as one single person and arrange all the "people" in one line, then arrange the couple. This technique is commonly referred to as "grouping". This time there are two such groups, so you can think that instead of having $8$ people, you have $5$ (three single people, one "couple" person and one "family" person) and then arrange the couple and the family.


Let us know if this hints are good or if you need extra help.