Illinois State University Mathematics Department

MAT 305: Combinatorics Topics for K-8 Teachers

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Quiz #01

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In the Braille system, a symbol, such as a lowercase letter, a punctuation mark, a suffix, and so on, is represented by raising at least one of the dots in the six-dot arrangement shown in figure (a), where the six Braille positions are labeled as well. For example, in figure (b), the dots in positions 1, 3, and 4 are raised and this six-dot arrangement represents the letter m. The definite article the is shown in (c), and the semicolon (;) is given by the six-dot arrangement in (d), where the dots in positions 2 and 3 are raised.

1. How many different symbols can be represented in the Braille system as described here?

Every dot can be either raised or not; therefore, for each of the 6 dots there are two choices. This results in 2*2*2*2*2*2=2^6=64 different patterns of raised/unraised dots. But, according to the Braille description, we must eliminate one of these 64 possible patterns, because at least one dot must be raised. Therefore, there are 63 possible raised-dot patterns.

2. How many symbols have an even number of raised dots?

An even number means there are 2, 4, or 6 raised dots. We use combinations to determine how many of each of these types exist, for combinations do not take into accord the order I choose the rasied dots, only the selection of dots from among the 6. There are C(6,2)+C(6,4)+C(6,6)=15+15+1=31 different raised-dot patterns that contain an even number of raised dots.

3. How many symbols have no more than one raised dot in each row, where dots 1 and 4 are in the same row, dots 2 and 5 are in the same row, and dots 3 and 6 are in the same row?

Consider the possible ways that this could happen. (a) Each row could have exactly one raised dot, (b) two of three rows could each have exactly one raised dot and the other row no raised dots, or (c) one of three rows could have exactly one raised dot and the other two each have no raised dots.

There is one way to choose the three rows to fit (a), and in each of the three rows there are two choices for which dot to raise. This yields 2*2*2=8 ways to have exactly one raised dot in each of three rows.

There are three ways to choose two of three rows to each have one raised dot, and within each of the two chosen rows there are two ways to choose a raised dot. Therefore, there are 3*2*2=12 ways to have no more than one raised dot per row when only two rows can have raised dots.

There are three ways to choose one of three rows to have one raised dot, and within the chosen row there are two ways to choose a raised dot. Therefore, there are 3*2=6 ways to have no more than one raised dot per row when only one rows can have raised dots.

We add the results because these are disjoint cases, so there are 8+12+6=26 ways to create a Braille symbol with no more than one raised dot per row.


A group of five judges listens to arguments and independently responds in one of three ways:

Agree, Disagree, No Judgment

The judges report their responses to an argument as an aggregate response such as 5&endash;0-0 (5 agree, 0 disagree, 0 no judgment) and 1-2-2 (1 agree, 2 disagree, 2 no judgment). The aggregate response is always in the order agree, disagree, no judgment.

4. Based on the individual judges' responses, how many ways are there for an aggregate response to come out as 2-3-0?

From among the 5 judges, choose two who find they agree with the argument. From the remaining 3 judges, select 3 who disagree with the argument. We use combinations here, because we are not concerned with the order by which the judgements are proclained. Therefore, there are C(5,2)*C(3,3)=10*1=10 ways for a 2-3-0 judgement to be rendered.

5. Based on the individual judges' responses, how many ways are there for an aggregate response to come out as a-b-0, where a and b represent all possible configurations for this context.

We follow the same logic as described in Question (4), so there are C(5,a)*C(b,b)=C(5,a) ways for a a-b-0 judgement to be rendered. We assume that a+b=5 here.


6. Sammy's sock drawer contains 12 identical black socks and 12 identical white socks. In the middle of the night, with no lights on, Sammy reaches into the drawer in search of two pairs of the same color socks. How many socks must Sammy pull from the drawer in order to be sure he has two pairs of the same color socks?

Sammy seeks four socks of the same color. The worst-case scenario is that Sammy pulls out six socks and gets three of each color. On his 7th pick, however, he must get a black or a white and therefore will have a group of 4 socks of the same color. By the pigeonhole principle, 7 socks are needed to assure us that Sammy has two pair of the same color.



Evaluation Criteria

  • Impact on Course Grade: Approximately 3%
  • Total Points: 30 points
  • There are 5 points available for each question. Approximately 3 points of each question is for the accuracy of your solution and 2 points on each question are for the clarity and completeness of your explanation.
  • In communicating your solutions, please describe any conditions or assumptions you made as you explored and solved the problems.


Administered 10 February 2003


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