10/12/2015MATH 106, Section 141 Section 14 Linear Equations with Unit Coefficients Questions about homework? Submit homework!

10/12/2015MATH 106, Section 142 Consider the equation x + y + z = 5. How many different solutions in non-negative integers are there? From Section 12, we know that the number of solutions is (n + 1)(n + 2) —————– = 2 (5 + 1)(5 + 2) —————– = 21 2 Let’s pretend we do not know the answer, and we take the following approach to solving this problem: We have 5 identical candy bars to be distributed to Xlyne, Yolanda, and Zelda. We shall denote the candy bars asc c c c c We shall use two slashes | | to denote three sections, the left one for Xlyne, the middle one for Yolanda, and the right one for Zelda. #1

10/12/2015MATH 106, Section 143 We have 5 identical candy bars to be distributed to Xlyne, Yolanda, and Zelda. We shall denote the candy bars asc c c c c We shall use two slashes | | to denote three sections, the left one for Xlyne, the middle one for Yolanda, and the right one for Zelda. For instance,c | c c | c cindicates that the number of candy bars Xlyne gets is, the number of candy bars Yolanda gets is, and the number of candy bars Zelda gets is As another example,c c c | | c cindicates that the number of candy bars Xlyne gets is, the number of candy bars Yolanda gets is, and the number of candy bars Zelda gets is

10/12/2015MATH 106, Section 144 The number of solutions to x + y + z = 5 in non-negative integers is the number of ways we can arrange the 7 symbols consisting of 5 “c” and 2 “|”. This number is 7! ——– = 21 2! 5! n = 5 2 is 1 less than the number of variables (people to whom candy bars are distributed) 7 is of course the sum of 2 and 5 C(7,2) = We can also think of this problem as choosing 2 positions for the slashes from the 7 total positions available in the arrangement of 5 candy bars and 2 slashes.

10/12/2015MATH 106, Section 145 IMPORTANT: The textbook uses the symbol “b” instead of slash “|” in its discussion. The textbook also uses the symbol “u” to represent the items being distributed (such as candy bars). Notice how this will affect the answer to homework problem #1.

10/12/2015MATH 106, Section 146 Consider the equation x 1 + x 2 + x 3 + x 4 = 5. How many different solutions in non-negative integers are there? 8! ——– = 56 3! 5! n = 5 3 is 1 less than the number of variables 8 is of course the sum of 5 and 3 = C(8,3) We can also think of this problem as choosing 3 positions for the slashes from the 8 total positions available in the arrangement of 5 candy bars and 3 slashes. #2

10/12/2015MATH 106, Section 147 Consider the equation x 1 + x 2 + … + x k = n. How many different solutions in non-negative integers are there? (n + k – 1)! ————— or C(n + k – 1, k – 1) (k – 1)! n! #3

10/12/2015MATH 106, Section 148 Find the number of different solutions in non-negative integers to each of the equations listed. u + v + x + y + z = 8 a + b + c + d + e + f = 12 x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 = 20 x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 = 6 u + v + w = 0 12! ——– = 495 4! 8! 17! ——– = ! 12! 26! ——– = ! 20! 13! ——– = ! 6! 1 #4

10/12/2015MATH 106, Section 149 A man has $100 that he intends to distribute as gifts to his 4 nephews and 6 nieces, all of different ages. How any ways can he distribute the money if he has 20 five-dollar bills for the 10 children with the possibility that one or more children gets nothing? he has 5 twenty-dollar bills for the 10 children, with the possibility that one or more children gets nothing? he has 20 five-dollar bills for the nephews only, with the possibility that one or more nephew gets nothing? #5 For each part of this problem, write an equation for which we want the number of solutions. Then we will find the number of solutions to each equation here in class; complete this exercise for next class, if we do not finish it today.

10/12/2015MATH 106, Section 1410 A man has $100 that he intends to distribute as gifts to his 4 nephews and 6 nieces, all of different ages. How any ways can he distribute the money if he has 20 five-dollar bills for the 10 children with the possibility that one or more children gets nothing? he has 5 twenty-dollar bills for the 10 children, with the possibility that one or more children gets nothing? he has 20 five-dollar bills for the nephews only, with the possibility that one or more nephews gets nothing? 29! ——– = 10,015,005 9! 20! 14! ——– = ! 5! 23! ——– = ! 20! x 1 + x 2 + … + x 9 + x 10 = 20 non-negative integers x 1 + x 2 + … + x 9 + x 10 = 5 non-negative integers x 1 + x 2 + x 3 + x 4 = 20 non-negative integers #5

10/12/2015MATH 106, Section 1411 he has 20 five-dollar bills for the nieces only, with the possibility that one or more nieces gets nothing? he has 5 twenty-dollar bills for the nephews only, with the possibility that one or more nephews gets nothing? he has 5 twenty-dollar bills for the nieces only, with the possibility that one or more nieces gets nothing? 8! ——– = 56 3! 5! 25! ——– = 53,130 5! 20! 10! ——– = 252 5! 5! x 1 + x 2 + x 3 + x 4 = 5 non-negative integers x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 20 non-negative integers x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 5 non-negative integers

10/12/2015MATH 106, Section 1412 Homework Hints: In Section 14 Homework Problem #12, just set up the calculation without actually calculating the final answer; you can think of assigning each ping pong ball to a color.

10/12/2015MATH 106, Section 1413 Consider the equation x + y + z = 5. How many different solutions in positive integers are there? From Section 12, we know that the number of solutions is (n – 1)(n – 2) —————– = 2 (5 – 1)(5 – 2) —————– = 6 2 Let’s pretend we do not know the answer, and we take the following approach to solving this problem: We have 5 identical candy bars to be distributed to Xlyne, Yolanda, and Zelda, where each girl is guaranteed to get at least one candy bar. Let us imagine that we remove 3 candy bars from the 5 that are to be distributed, for the purpose of giving each girl the one that is guaranteed. We have 5 – 3 = 2 identical candy bars to be distributed to the 3 girls. #6

10/12/2015MATH 106, Section 1414 We have 5 – 3 = 2 identical candy bars to be distributed to the 3 girls. The number of ways to do this is the same as the number of ways we can solve the equation x + y + z = 2 in the non-negative integers. n minus the number of variables (3) We now know that this number of solutions is 4! ——– = 6 2! 2! C(4,2) =

10/12/2015MATH 106, Section 1415 Consider the equation x 1 + x 2 + … + x k = n. What is the smallest integer n could be in order for solutions of the equation in positive integers to exist? How many different solutions in positive integers are there? n cannot be any smaller than k (the number of variables). Let us imagine that we remove 1 “item” for each variable, and recognize that the number of solutions in positive integers to x 1 + x 2 + … + x k = n is the same as the number of solutions in non-negative integers to x 1 + x 2 + … + x k = n – k, that is, (n – 1)! —————— or C(n – 1, k – 1) (k – 1)! (n – k)! #7

10/12/2015MATH 106, Section 1416 A man has $100 that he intends to distribute as gifts to his 4 nephews and 6 nieces, all of different ages. How any ways can he distribute the money if he has 20 five-dollar bills for the 10 children with the guarantee that each child gets at least one bill? he has 5 twenty-dollar bills for the 10 children, with the guarantee that each child gets at least one bill? he has 20 five-dollar bills for the nephews only, with the guarantee that each nephew gets at least one bill? 19! ——– = 9! 10! 92,378 This is not possible. 19! ——– = 969 3! 16! x 1 + x 2 + … + x 9 + x 10 = 20 positive integers x 1 + x 2 + … + x 9 + x 10 = 10 non-negative integers x 1 + x 2 + … + x 9 + x 10 = 5 positive integers x 1 + x 2 + … + x 9 + x 10 = –5 non-negative integers x 1 + x 2 + x 3 + x 4 = 20 positive integers x 1 + x 2 + x 3 + x 4 = 16 non-negative integers #8

10/12/2015MATH 106, Section 1417 he has 20 five-dollar bills for the nieces only, with the guarantee that each niece gets at least one bill? he has 5 twenty-dollar bills for the nephews only, with the guarantee that each nephew gets at least one bill? he has 5 twenty-dollar bills for the nieces only, with the guarantee that each niece gets at least one bill? 4! ——– = 4 3! 1! 19! ——– = 5! 14! 11,628 This is not possible. x 1 + x 2 + x 3 + x 4 = 5 positive integers x 1 + x 2 + x 3 + x 4 = 1 non-negative integers x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 20 positive integers x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 14 non-negative integers x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 5 positive integers x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = –1 non-negative integers

10/12/2015MATH 106, Section 1418 Each of 15 identical bonuses is to be distributed among all the different employees in a particular office. How many ways are there to do this, if there are 20 people in the office, and there is no restriction on the number of bonuses one employee can get? How many ways are there to do this, if there are 10 people in the office, and there is no restriction on the number of bonuses one employee can get? How many ways are there to do this, if there are 20 people in the office, and each employee is guaranteed to get at least one bonus? #9 Complete all the exercises on this handout as part of the homework for next class, if we do not finish them today in class.

10/12/2015MATH 106, Section 1419 Each of 15 identical bonuses is to be distributed among all the different employees in a particular office. How many ways are there to do this, if there are 20 people in the office, and there is no restriction on the number of bonuses one employee can get? How many ways are there to do this, if there are 10 people in the office, and there is no restriction on the number of bonuses one employee can get? How many ways are there to do this, if there are 20 people in the office, and each employee is guaranteed to get at least one bonus? 34! ——–– = 1,855,967,520 19! 15! 24! ——– = 1,307,504 9! 15! x 1 + x 2 + … + x 19 + x 20 = 15 non-negative integers x 1 + x 2 + … + x 9 + x 10 = 15 non-negative integers #9

10/12/2015MATH 106, Section 1420 Each of 15 identical bonuses is to be distributed among all the different employees in a particular office. How many ways are there to do this, if there are 20 people in the office, and there is no restriction on the number of bonuses one employee can get? How many ways are there to do this, if there are 10 people in the office, and there is no restriction on the number of bonuses one employee can get? How many ways are there to do this, if there are 20 people in the office, and each employee is guaranteed to get at least one bonus? 34! ——–– = 1,855,967,520 19! 15! 24! ——– = 1,307,504 9! 15! This is not possible. x 1 + x 2 + … + x 9 + x 20 = 15 non-negative integers x 1 + x 2 + … + x 9 + x 10 = 15 non-negative integers x 1 + x 2 + … + x 9 + x 20 = 15 positive integers x 1 + x 2 + … + x 9 + x 20 = –5 non-negative integers

10/12/2015MATH 106, Section 1421 How many ways are there to do this, if there are 10 people in the office, and each employee is guaranteed to get at least one bonus? How many ways are there to do this, if there are 20 people in the office, and each employee can get at most one bonus? Each of 15 different bonuses (one is $15, one is $14, one is $13, etc.) is to be distributed among all the different employees in a particular office. How many ways are there to do this, if there are 20 people in the office, and each employee can get at most one bonus? 14! ——– = ! 5! From the total of 20 people, we must choose the 15 people who will each get one bonus: C(20,15) = 15,504 P(20,15) =  x 1 + x 2 + … + x 9 + x 10 = 15 positive integers x 1 + x 2 + … + x 9 + x 10 = 5 non-negative integers #10

10/12/2015MATH 106, Section 1422 Homework Hints: In Section 14 Homework Problem #10, In Section 14 Homework Problem #11, notice that you are simply choosing four coins from five. let p be the number of pennies in the collection, let n be the number of nickels in the collection, let d be the number of dimes in the collection, etc. Quiz #3 NEXT CLASS! Be sure to do the review problems for this, quiz posted on the internet. The link can be found in the course schedule.