Illinois State University Mathematics Department
MAT 305: Combinatorics Topics for K8 Teachers 
Sample Responses to Course Assignments 


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The smallest of r consecutive integers is n. What is the largest? 
How many integers between 1 and 2000 are multiples of 11 but not multiples of 3? 
In a group of 5 people, show that there are two who have the same number of acquaintances in the group. Generalize your result to show that in a group of n people, there are two who have the same number of acquaintances. 



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n + r  1 is the largest 
181/3 = 60 
If a and b know each other and no one else in the group, they have the same number of acquaintances. It doesn't matter whether the group has 5 people or n people. 
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n + r  1 represents largest integer; 
121 
In a group of 5 people, there could be 04 acquaintances
[chart]: 

n = smallest integer 
121: delete 33,66,99, . . . , 1980 = 60 
If one person knows 0 then have four people to fill 3 boxes. If everyone knows at least one person, then 4 boxes to fill. 



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n + r  1 
11,22,33,44,55,66,77,88,99 [Solver showed large X
through 33 and 66 and 99.] 
There are five possible combinations for five people, so
it seems that there could always be a different one for each
person. However, since it is impossible for one person in
the group to be an acquaintance with everyone else (4,0)
while another person in the group has no acquaintances
(0,4), they are mutually exclusive. since that leaves only
four combinations for five people, two people will have the
same combination of acquaintances. 
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r = 8, n = 50, 50 + 8 = 58, n + r  1 
33,66,99, . . . , 1980: multiples of 3 and 11 
With 5 people, the number of acquaintances is only 04.
You can never have 04 inclusive because one person cannot
know anyone and someone else know everyone. Therefore, one
number has to be duplicated. Then there are at least two
people with the same number of acquaintances. 



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n + r  1 
There are 181 multiples of 11 from 1 to 2000. The largest
common multiple of 3 and 11 (i.e., 33) in the sequence from
1 to 2000 is 1980: 33, . . . , 1980. Then using the result
shown in problem 8, (198033)/33 + 1 = 60. We also could
have generated the value 60 by noting that 2000/33 = 60
remainder 20. 
If no one knows anyone else in the group, they all have 0
acquaintances and we've met the condition. Likewise, if all
five people know each other, they each have 4
acquaintances. 
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n + r  1 is the largest 




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In the expansion of the binomial for any term state the restrictions on a and b. 
At Bagwanna State University, students' programs of study must include course credit in mathematics, science, English, history, and the arts. Each student must complete a total of exactly 46 credits that include these areas of study. Furthermore, no less than 6 credits must be in mathematics, at least 8 credits must be in science, and 10 or more credits must be from English. How many different student programs, based on credits earned, can be designed under these restrictions? 



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(2 of 4 points earned) The sum of a + b must be equal to T. a + b = t 
(10 of 15 points earned) >= 6 math 24 determined 
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(2 of 4 points earned) a + b = t 
(10 of 15 points earned) C(27,5) How many ways can I design a program with the 22 remaining credits and the six categories? I have 22 credits and 5 borders to place, thus I have 27 items and I must choose 5 places to place my borders. C(27,5) represents the number of different student programs that can be designed under the given restrictions. 



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(3 of 4 points earned) (n + m)(n + m)(n + m) . . . a + b = t a >= 0 but a & b are both = 0 if and only if t = 0 
(12 of 15 points earned) C(26,4) or 14,950 10 + 8 + 6 = 24 accounted for; 46  24 = 22 left to determine: C(26,4) Because 24 hours have been accounted for there are 22 left to determine. I combined these 22 hours with the 4 dividers and chose the placement of the four dividers to determine the number of hours for each area of study. There are C(26,4) possible different student programs. 
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(3 of 4 points earned) The sum of a and b must equal t. A or B may not be larger than t. 
(12 of 15 points earned) C(29,4) We now have 25 credits for the 5 subjects. Based on this we must solve over the positive integers, since each subject needs at least one. Using C(m + r  1, r  1) where "m" equals the desired responses (25) and r  1 is the number of dividers needed to separate the 5 subject areas we conclude that C(25 + 5  1, 5  1) or C(29,4) represents the number of different student programs to design. 



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(4 of 4 points earned) a + b = t Note: 
(15 of 15 points earned) C(24,4) M >= 6, S >= 8, E >= 10, H >= 1, A >=
1 Now able to select over nonnegative numbers C(n + k 1, k  1), 20 + 5 1, 5  1 C(24,4) 
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(4 of 4 points earned)
where J represents the set of integers 
(15 of 15 points earned) C(24,4) M >= 6 (M represents mathematics) I am assuming there are 1credit courses in history and the arts because it is not specified and a student must have course credit in all 5. M + S + E + H + A = 46, but 26 credits are accounted for by restrictions so (46  26 = 20) M + S + E + H + A = 20. This equation can be solved over the nonnegative integers now. C(24,4) (There are 5 categories with 4 dividers and 46 tally marks must be placed with the placement of some prespecified.) 

