Illinois State University Mathematics Department

MAT 409: Topics in Algebra and Combinatorics for K-8 Teachers





Notes on Assignment #6



  • Chapter 2 Review (p. 35-36): 3
  • Chapter 3 Section 3 (p. 76): 30
  • Chapter 3 Section 4 (pp. 94-98): 1,3,4,7,8,9,11,13,15,17,19,23,25,29,33,35,37,38



Chapter 2 Review
  • #3: (a) To have a pair, Tim needs a left-hand glove and a right-hand glove. There are 12 of each in the drawer. The worst case scenario is that Tim first picks all 12 gloves that all are for one hand. When he chooses a 13th glove, however, he's assured of a pair. In fact, the response is the same for (b), because, if Tim has picked all 12 gloves for the same hand, he'll have every color. The 13th pick assures that there will be a color match as well as a left/right pair.
Chapter 3 Section 3
  • #30:
    • Situation I: Assume that order or arrangement is significant. For example, a white hat on Tim and a green hat on Tom is different from a green hat on Tim and a white hat on Tom.
      • (a) Each brother has 7 choices, so the number of 5-hat sets is 7󬱟󬱟 = 75.
      • (b) P(7,5): The first brother has 7 colors to choose from, the next has 6, and so on, down to the last brother who has 3 colors to choose from.
      • (c) There are 7 colors for the hats. Therefore, there are 7 ways the set of four brothers could have chosen the same color hat. There remain 6 colors for the fifth brother to choose. Finally, there are 5 ways to choose the one brother with a hat a different color from the other. Therefore, this could have occurred in 7󬝱 = 210 ways.
    • Situation II: Assume that order or arrangement is not significant. For example, a white hat on Tim and a green hat on Tom is the same as a green hat on Tim and a white hat on Tom, because, from a hat perspective only, we have a white hat and a green hat.
      • (a) We need to determine the possibilities for hat color combinations, and with each of those determine the number of ways it could occur. This includes: (i) All hats are the same color. There are 7 colors, so there are 7 ways to have all 5 hats the same color. (ii) Four (4) hats the same color and 1 hat that is a different color. There are 7 ways to choose a color for the four like-colored hats and 6 ways to choose a color for the fifth hat, so 42 in all. (iii) Three (3) hats the same color and the other two hats match in a different color. There are 7 ways to choose a color for the three like-colored hats and 6 ways to choose a color for the two hats that match a different color, so 42 in all. (iv) Three (3) hats the same color and the other two hats do not match each other. There are 7 ways to choose a color for the three like-colored hats, and C(6,2) for the other two, so 715 = 105 in all. (v) Two (2) hats match color, two other hats match color, and the fifth hat differs from the first two colors. There are C(7,2)5 = 105 in all for this option. (vi) Two hats match in color and the other three hats are different from that and different from each other. There are 7證(6,3) = 140 in all. (vii) All five hats differ in color. There are C(7,5) = 21 ways for this to occur. This yields a total of 462 options.
      • (b) C(7,5)
      • (c) There are 7 colors for the hats. Therefore, there are 7 ways the set of four brothers could have chosen the same color hat. There remain 6 colors for the fifth brother to choose. Therefore, this could have occurred in 76 = 42 ways.
Chapter 3 Section 4
  • #1: C(20,3) = 20!/(17!3!) = 1140
  • #3: C(12,6) = C(11,5) + C(11,6)
  • #4: We know C(21,8) = C(20,7) + C(20,8), so C(21,8) - C(20,7) = C(20,8).
  • #7: Using Pascal's Formula, C(n,k) + C(n,k+1) = C(n+1,k+1).
  • #8: (r + s)3 = C(3,0)r3 + C(3,1)r2s + C(3,2)rs2 + C(3,3)s3 = r3 + 3r2s + 3rs2 + s3
  • #9: Because we have n repetitions of the factor (x + y), the collected-terms expansion will have terms that include from n repetitions of x to 0 repetitions of x. This means there are n+1 collected terms.
  • #11: (a) 212 uncollected terms (b) When a occurs 5 times, we know (2a)5 is a factor, as is (-b)7. We also know there are C(12,5) = C(12,7) ways to have this pair of factors occur. Therefore, the collected term is C(12,5)*(2a)5(-b)7= 792*32a5*(-1)b7 = -25344a5b7. This shows that H = -25,344.
  • #13: (a) There are 412 uncollected terms. (b) The value of J is (12!)/(5!1!0!6!) = 5544.
  • #15: There are 5 + 1 = 6 collected terms.
  • #17: (a) 410 uncollected terms (b) The value of K is (10!)/(3!1!2!4!) = 12,600.
  • #19: C(7,0)*(4m)7(3t)0 + C(7,1)*(4m)6(3t)1 + C(7,2)*(4m)5(3t)2= 1*47*m7*30*t0 + 7*46*m6*31*t1 + 21*45*m5*32*t2 =  16.384m7 + 86,016m6t + 193,536m5t2
  • #23: (a) K = C(10,7) = C(10,3) = 120 (b) C(10,0) + C(10,1) + ... + C(10,9) + C(10,10) = 210
  • #25: No. The exponents in the term do not sum to 12, and every term in that expansion must have exponents that sum to 12.
  • #29: (a) 511 (b) K = 11!/(4!2!1!4!) = 34,650 (c) The only way to get 11 as a coefficient is for the multinomial coefficient 11!/(e1!e2!e3!e4!e5!) = 11, where e1+ e2+ e3+ e4+ e5 = 11 and each ei is a non-negative integer no greater than 11. This can only occur when the denominator product is 10!, because 11!/(k) = 11 implies k = 10!. The only way for the denominator to be 10!, with the restrictions on ei already stated, is for one of the ei to be 10, another to be 1, and the other three to be 0. This can be done in 20 ways, because there are 5 ways to select the first ei, 4 ways to select the next, and the remaining three must then be used.
  • #33: Solutions will vary.
  • #35: (a) Here, for each of Seth's card selections, he has 10 choices. Therefore, there are 105 possible ways for Seth to make his selection. (b) Now Seth can choose no more than one of a kind, and the order of selection is not a factor. Therefore, there are C(10,5) selections possible.
  • #37: We know S(k) = 2k. This gives us S(12) - S(10) = 212 - 210 = 210(22 - 20) = 210(4 - 1) = 210(3) = 3*S(10).
  • #38: Solution: 2[C(9,9)+C(10,9)+C(11,9)++C(18,9)] Think about this as a modification or application of a grid problem. A "path" passes through integer ordered pairs (x,y), starting at the origin (0,0) and ending at either (a) an ordered pair on the line y = 10, with x an integer ranging from 0 to 9, or (b) an ordered pair on the line x = 10, with y an integer ranging from 0 to 9. We just need to count the paths, using the grid-problem strategy, required to get to each of these ending points. The only way to get to (x,10) or (10,y), for x <= 9 and y <=9, is to get there from (x,9) or from (9,y), respectively. That is, once the value "10" is reached, there is no more moving along the horizontal line y=10 or the vertical line x=10. As soon as a team scores 10, the match is over. The picture below may help illustrate this.
grid for #38