1. 1/23/2016MATH 106, Section 171 Section 17 More Equations with Restricted Solutions Questions about homework? Submit homework!

2. 1/23/2016MATH 106, Section 172 Before beginning Section 17: Recall the Inclusion/Exclusion formulas involving the union of two sets and the union of three sets: Then, look at the formula involving the union of four sets, derived on page 118 of the textbook. #(A  B) = #A + #B – #(A  B) #(A  B  C) = #A + #B + #C – #(A  B) – #(A  C)– #(B  C)+ #(A  B  C) #(A  B  C) = #A + #B + #C + #D – #(A  B) – #(A  C)– #(A  D)– #(B  C)– #(B  D) – #(C  D)

3. 1/23/2016MATH 106, Section 173 Then, look at the formula involving the union of four sets, derived on page 118 of the textbook. #(A  B  C  D) = #A + #B + #C + #D – #(A  B) – #(A  C)– #(A  D)– #(B  C)– #(B  D) – #(C  D) + #(A  B  C) – #(A  B  D)– #(A  C  D)– #(B  C  D) – #(A  B  C  D) How can we be sure that we have correctly obtained every pair of sets? We know that the number of pairs must be C(4,2) = 6. How can we be sure that we have correctly obtained every triple of sets? We know that the number of triples must be C(4,3) = 4.

4. 1/23/2016MATH 106, Section 174 Use the pattern observed with this formula when doing Exercise #2 in the Section 16 Homework. #(A  B  C  D  E) = #A + #B + #C + #D + #E – etc. How can you be sure that you have correctly obtained every pair of sets? We know that the number of pairs must be C(5,2) = 10. How can you be sure that you have correctly obtained every triple of sets? etc. We now begin the Section #17 Handout. We will work on #1 here in class and finish whatever we do not get to today in the next class. You will do #2 and the short essay which follows #2 as part of the homework for next class.

5. 1/23/2016MATH 106, Section 175 Consider the equation v + w + x + y + z = 14. How many different solutions in non-negative integers are there? How many different solutions in non-negative integers are there, if x must be less than or equal to 3? (n + k – 1)! We know this is C(n + k – 1, k – 1) = ————— = (k – 1)! n! 18! ——– = 3060 4! 14! v + w + x + y + z = 14 x  3 v + w + x + y + z = 14 4  x We shall use GOOD = ALL – BAD. 14! ——– = 1001 4! 10! v + w + x + y + z = 10 non-negative integers The desired number of solutions is 3060 – 1001 = 2059 (a) (b) #1

6. 1/23/2016MATH 106, Section 176 How many different solutions in non-negative integers are there, if x must be less than or equal to 3, and y must be less than or equal to 2? v + w + x + y + z = 14 x  3 AND y  2 v + w + x + y + z = 14 4  x OR 3  y We shall use GOOD = ALL – BAD. #[(4  x)  (3  y)] = #(4  x) + #(3  y) – #[(4  x)  (3  y)] v + w + x + y + z = 14 4  x v + w + x + y + z = 14 3  y v + w + x + y + z = 14 4  x AND 3  y 14! ——– = 1001 4! 10! 15! ——– = 1365 4! 11! 11! ——– = 330 4! 7! 1001 + 1365 – 330 = 2036 solutions The number of solutions is 3060 – 2036 = 1024 (c)

7. 1/23/2016MATH 106, Section 177 How many different solutions in non-negative integers are there, if we must have x  2, y  3, and z  4 ? v + w + x + y + z = 14 x  2 AND y  3 AND z  4 v + w + x + y + z = 14 3  x OR 4  y OR 5  z We shall use GOOD = ALL – BAD. #[(3  x)  (4  y)  (5  z)] = #(3  x) + #(4  y) + #(5  z) – #[(3  x)  (4  y)] – #[(3  x)  (5  z)] – #[(4  y)  (5  z)] + #[(3  x)  (4  y)  (5  z)] 15! ——– + 4! 11! 1365 + 14! ——– + 4! 10! 1001 + 13! ——– – 4! 9! 715 – 11! ——– – 4! 7! 330 – 10! ——– – 4! 6! 210 – 9! ——– + 4! 5! 126 + 6! ——– = 4! 2! 15 = 2430 The number of solutions is 3060 – 2430 = 630 (d)

8. 1/23/2016MATH 106, Section 178 How many different solutions in non-negative integers are there, if x, y, and z must each be less than or equal to 2? v + w + x + y + z = 14 x  2 AND y  2 AND z  2 v + w + x + y + z = 14 3  x OR 3  y OR 3  z We shall use GOOD = ALL – BAD. #[(3  x)  (3  y)  (3  z)] = #(3  x) + #(3  y) + #(3  z) – #[(3  x)  (3  y)] – #[(3  x)  (3  z)] – #[(3  y)  (3  z)] + #[(3  x)  (3  y)  (3  z)] 15! ——– + 4! 11! 1365 + 15! ——– + 4! 11! 1365 + 15! ——– – 4! 11! 1365 – 12! ——– – 4! 8! 495 – 12! ——– – 4! 8! 495 – 12! ——– + 4! 8! 495 + 9! ——– = 4! 5! 126 = 2736 The number of solutions is 3060 – 2736 = 324 (e)

9. Complete as much of the Section #17 Handout as possible before you leave class today, and also set up the equations for each of the two parts of #2 on the Section #17 Handout. 1/23/2016MATH 106, Section 179 For next class, do Problem #8 in the Section 16 Homework: In Problem #8, use #U – #(A  B) = #U – [#A + #B – #(A  B)] perfect squares 1 2 = 2 2 = 1 4 3 2 = 9 How many of these are there? ( ) 2 = 1,000,000 perfect cubes 1 3 = 2 3 = 1 8 3 3 = 27 How many of these are there? ( ) 3 = 1,000,000 How many are both a perfect square and a perfect cube? There are also some Section 17 homework problems

10. Homework Hints: In Section 17 Homework Problems #1, #2, #3, #4, and #5, the approach used in part (c) of #1 on the Section #17 Handout can be used. 1/23/2016MATH 106, Section 1710

11. 1/23/2016MATH 106, Section 1711 How many different solutions in non-negative integers are there, if w, x, y, and z must each be less than or equal to 2? v + w + x + y + z = 14 w  2 AND x  2 AND y  2 AND z  2 v + w + x + y + z = 14 3  w OR 3  x OR 3  y OR 3  z We shall use GOOD = ALL – BAD. #[(3  w)  (3  x)  (3  y)  (3  z)] = #(3  w) + #(3  x) + #(3  y) + #(3  z) – #[(3  w)  (3  x)] – … – #[(3  y)  (3  z)] + #[(3  w)  (3  x)  (3  y)] + … + #[(3  x)  (3  y)  (3  z)] – #[(3  w)  (3  x)  (3  y)  (3  z)] 15! (4)——– – 4! 11! 12! ——– + 4! 8! C(4,2) 9! ——– – 4! 5! C(4,3) 6! ——– = 4! 2! 5460 – 2970 + 504 – 15 = 2979 number of solutions = 3060 – 2979 = 81 (f)

12. 1/23/2016MATH 106, Section 1712 How many different solutions in non-negative integers are there, if w, x, y, and z must each be less than or equal to 3? v + w + x + y + z = 14 w  3 AND x  3 AND y  3 AND z  3 v + w + x + y + z = 14 4  w OR 4  x OR 4  y OR 4  z We shall use GOOD = ALL – BAD. #[(4  w)  (4  x)  (4  y)  (4  z)] = #(4  w) + #(4  x) + #(4  y) + #(4  z) – #[(4  w)  (4  x)] – … – #[(4  y)  (4  z)] + #[(4  w)  (4  x)  (4  y)] + … + #[(4  x)  (4  y)  (4  z)] – #[(4  w)  (4  x)  (4  y)  (4  z)] 14! (4)——– – 4! 10! 10! ——– + 4! 6! C(4,2) 6! ——– – 4! 2! C(4,3) 4004 – 1260 + 60 = 2804 number of solutions = 3060 – 2804 = 256 0 = (g)

13. 1/23/2016MATH 106, Section 1713 How many different solutions in non-negative integers are there, if every variable must each be less than or equal to 3? v + w + x + y + z = 14 v  3 AND w  3 AND x  3 AND y  3 AND z  3 v + w + x + y + z = 14 4  v OR 4  w OR 4  x OR 4  y OR 4  z We shall use GOOD = ALL – BAD. #[(4  v)  (4  w)  (4  x)  (4  y)  (4  z)] = #(4  v) + #(4  w) + #(4  x) + #(4  y) + #(4  z) – #[(4  v)  (4  w)] – … – #[(4  y)  (4  z)] + #[(4  v)  (4  w)  (4  x)] + … + #[(4  x)  (4  y)  (4  z)] – #[(4  v)  (4  w)  (4  x)  (4  y)] – … – #[(4  w)  (4  x)  (4  y)  (4  z)] + #[(4  v)  (4  w)  (4  x)  (4  y)  (4  z)] (h)

14. 1/23/2016MATH 106, Section 1714 How many different solutions in non-negative integers are there, if every variable must each be less than or equal to 3? #[(4  v)  (4  w)  (4  x)  (4  y)  (4  z)] = #(4  v) + #(4  w) + #(4  x) + #(4  y) + #(4  z) – #[(4  v)  (4  w)] – … – #[(4  y)  (4  z)] + #[(4  v)  (4  w)  (4  x)] + … + #[(4  x)  (4  y)  (4  z)] – #[(4  v)  (4  w)  (4  x)  (4  y)] – … – #[(4  w)  (4  x)  (4  y)  (4  z)] + #[(4  v)  (4  w)  (4  x)  (4  y)  (4  z)] 14! (5)——– – 4! 10! 10! ——– + 4! 6! C(5,2) 6! ——– – 4! 2! C(5,3) 5005 – 2100 + 150 = 3055 0 + 0 = (h)

15. 1/23/2016MATH 106, Section 1715 How many different solutions in non-negative integers are there, if every variable must each be less than or equal to 3? v + w + x + y + z = 14 v  3 AND w  3 AND x  3 AND y  3 AND z  3 v + w + x + y + z = 14 4  v OR 4  w OR 4  x OR 4  y OR 4  z We shall use GOOD = ALL – BAD. number of solutions = 3060 – 3055 = 5 14! (5)——– – 4! 10! 10! ——– + 4! 6! C(5,2) 6! ——– – 4! 2! C(5,3) 5005 – 2100 + 150 = 3055 0 + 0 = only 5 solutions?!?!?!?!? We should be able to list them. v = w = x = y = 3, z =2 v = w = x = z = 3, y =2 v = w = y = z = 3, x =2 v = x = y = z = 3, w =2 w = x = y = z = 3, v =2 (h)

16. 1/23/2016MATH 106, Section 1716 How many different solutions in non-negative integers are there, if every variable must each be less than or equal to 2? DON’T DO ANY WORK FOR THIS PROBLEM!!!! Think about it!! The number of solutions is zero (0). Set up the equations for each of the two parts of the next exercise before you leave class today, and for homework next class: Complete the Section #17 Handout to be submitted as part of a future homework assignment. (i)

17. 1/23/2016MATH 106, Section 1717 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 (a) he has 20 five-dollar bills for the 10 children with the only restriction that the three children under 10 years old get at most 10 dollars each? x 1 + x 2 + … + x 9 + x 10 = 20 x 8  2 AND x 9  2 AND x 10  2 x 1 + x 2 + … + x 9 + x 10 = 20 3  x 8 OR 3  x 9 OR 3  x 10 We shall use GOOD = ALL – BAD. #[(3  x 8 )  (3  x 9 )  (3  x 10 )] = #(3  x 8 ) + #(3  x 9 ) + #(3  x 10 ) – #[(3  x 8 )  (3  x 9 )] – #[(3  x 8 )  (3  x 10 )] – #[(3  x 9 )  (3  x 10 )] + #[(3  x 8 )  (3  x 9 )  (3  x 10 )] #2

18. 1/23/2016MATH 106, Section 1718 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 (a) he has 20 five-dollar bills for the 10 children with the only restriction that the three children under 10 years old get at most 10 dollars each? x 1 + x 2 + … + x 9 + x 10 = 20 x 8  2 AND x 9  2 AND x 10  2 x 1 + x 2 + … + x 9 + x 10 = 20 3  x 8 OR 3  x 9 OR 3  x 10 We shall use GOOD = ALL – BAD. 26! (3)——– – 9! 17! 23! (3)——– + 9! 14! 20! ——– = 9! 11! 9,373,650 – 2,451,570 + 167,960 = 7,090,040 The number of solutions is 10,015,005 – 7,090,040 = 2,924,965

19. 1/23/2016MATH 106, Section 1719 (b) he has 20 five-dollar bills for the 10 children with the restrictions that Sandy gets at most 10 dollars, Terri gets at most 15 dollars each, and Matt gets at most 20 dollars? x 1 + x 2 + … + x 9 + x 10 = 20 x 8  2 AND x 9  3 AND x 10  4 x 1 + x 2 + … + x 9 + x 10 = 20 3  x 8 OR 4  x 9 OR 5  x 10 We shall use GOOD = ALL – BAD. #[(3  x 8 )  (4  x 9 )  (5  x 10 )] = #(3  x 8 ) + #(4  x 9 ) + #(5  x 10 ) – #[(3  x 8 )  (4  x 9 )] – #[(3  x 8 )  (5  x 10 )] – #[(4  x 9 )  (5  x 10 )] + #[(3  x 8 )  (4  x 9 )  (5  x 10 )]

20. 1/23/2016MATH 106, Section 1720 (b) he has 20 five-dollar bills for the 10 children with the restrictions that Sandy gets at most 10 dollars, Terri gets at most 15 dollars each, and Matt gets at most 20 dollars? x 1 + x 2 + … + x 9 + x 10 = 20 x 8  2 AND x 9  3 AND x 10  4 x 1 + x 2 + … + x 9 + x 10 = 20 3  x 8 OR 4  x 9 OR 5  x 10 We shall use GOOD = ALL – BAD. 26! ——– + 9! 17! 25! ——– + 9! 16! 24! ——– – 9! 15! 22! ——– – 9! 13! 21! ——– – 9! 12! 20! ——– + 9! 11! 17! ——– = 9! 8! 3,124,550 + 2,042,975 + 1,307,504 – 497,420 – 293,930 – 167,960 + 24,310 = 5,540,029 The number of solutions is 10,015,005 – 5,540,029 = 4,474,976

21. 1/23/2016MATH 106, Section 1721 Check the date for Quiz #5 Be sure to do the review problems for this, quiz posted on the internet. The link can be found in the course schedule. Homework Hints: In Section 17 Homework Problem #6, In Section 17 Homework Problem #7, In Section 17 Homework Problem #8, the approach that can be used is one which combines the approaches used in parts (d) (f), and (g) of #1 on the Section #17 Handout. 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, and let q be the number of quarters in the collection. Then write the required equation.

22. 1/23/2016MATH 106, Section 1722 Homework Hints: In Section 17 Homework Problem #6, In Section 17 Homework Problem #7, In Section 17 Homework Problem #8, the approach that can be used is one which combines the approaches used in parts (d) (f), and (g) of #1 on the Section #17 Handout. 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, and let q be the number of quarters in the collection. Then write the required equation. use the equation from #7, and add the appropriate restrictions on the variables (i.e., notice how the number of pennies is restricted, the number of nickels is restricted, etc.).