AMS 301 Sect



AMS 301 Test 2 Spring 2021Find the coefficient of x28 in (x3+ . . . + x8)5. = [ x3(1+ x+ x2+. . x5)]5 = x15[(1 – x6)5/(1 – x)5] want coefficient of x28-15= x13 (1 – x6)5 = 1 – C(5,1)x6 + C(5,2)x12 + . . [1/(1-x)]5 = . . . C(k+5-1,k)xk a0b13 + a6b7 + a12b1 Coeff. of x13 = 1?C(13+5-1,13) - C(5,1)?C(7+5-1,7) + C(5,2)?C(1+5-1,1).2. Consider the problem of counting the ways to distribute 31 identical objects in 6 boxes with at least three objects in each box.a) Model this problem as an integer-solution-of-equation-problem. e1 + . + e6=31, ei ≥3b) Model this problem as a certain coefficient of a generating function. (x3+ . . . )6, coef of x31c) Solve this problem. C(6+(31-18)-1,(31-18))3. How many 5-letter sequences (formed from the 26 letters in the alphabet, with repetition allowed) contain exactly two A’s and exactly one N? ANS C(5,2)C(3,1)242 or C(5,1)C(4,2)242.4. Find the probability that an 11-card hand (from a 52-card deck) has exactly 4 pairs (no 4-of-a-kinds and no 3-of-a-kinds). ANS : Denominator C((52,11); Numerator: C(13,4) (pick which 4 of the 13 kinds have a pair)?C(4,2)4 (pick the 2 cards for each of these 4 pairs)?C(9,3)(pick which 3 kinds have 1-of-a-kind)?C(4,1)3 (pick the card for each of these 3 kinds)5. How many arrangements of 5 A’s, 6 B’s, and 8 C’s in which the first A occurs before the first B (the first B is not necessarily immediately after the first A)? ANS: C(19,11) (pick subset of positions for 5 A’s and 6 B’s) x 1 (put an A in first of these 11 positions) x 10!/4!6! (arrange other 4 A’s and 6’s in the other 10 positions for A’s and B’s) x 1 (place C’s in remaining 8 positions).6. How many arrangements of letters in DISAPPEARANCES have ALL of the following properties: (i) there are at least two letters between each A, (ii) begins with a consonant, and (iii) the consonants are not in alphabetical order. ANS: i) set aside 1st position for consonant- worry about other 13 positions ??????? ii) spacing A’s among other 10 letters (excluding one cons.) C(4+ (10-4)-1,(10-4))??? iii) pick 7 positions for other consonants (along with first position), C(10,7) ??? iv) arrange the 8 consonants in all non-alphabetical orders (all orders minus 1) (8!/2!2! – 1)??? v) ?arrange other vowels. ?3!/2! ANS: C(4+ (10-4)-1,(10-4)) x C(10,7) x (8!/2!2! -1) x 3!/2!7. How many integer solutions are there to 2x1 + 2x2 + 2x3 + x4 + x5 = 9 with xi ≥ 0? (Hint: break into cases) Let k be the sum of x1 + x2 + x3, k=0 or 1 or 2 or 3 or 4. Consider k=2. Then answer of this subcase is solutions to x1 + x2 + x3=2 – C(3+2-1,2) – times the solutions to x4 + x5 = 5 (=9-2x2) – C(2+5-1,5) Final answer sum (k=0,1,2,3,4) {C(3+k-1,k)xC(2+(9-2k)-1, (9-2k) )} AMS 301 Test 2 Fall 20201. Find the coefficient of x35 in (x2+ . . . + x8)7. = [ x2(1+ x+ x2+. . x6)]7 = x14[(1 – x7)7/(1 – x)7] want coefficient of x35-14= x21 (1 – x7)7 = 1 – C(7,1)x7 + C(7,2)x14 + . . [1/(1-x)]7 = . . . C(k+7-1,k)xk a0b21 + a7b14 + a14b7 + a21b0 Coeff. of x21 = 1?C(21+7-1,21) - C(7,1)?C(14+7-1,14) + C(7,2)?C(7+7-1,7) - C(7,3) ?1.2. Consider the problem of counting the ways to distribute 29 identical objects in 7 boxes with at least two objects in each box.a) (2 pts) Model this problem as an integer-solution-of-equation-problem. e1 + . + e7=29, ei >=2b) (3 pts) Model this problem as a certain coefficient of a generating function. (x2+ . . . )7, coef of x29c) (3 pts) Solve this problem. C(7+(29-14)-1,(29-14))3. How many sequences of length 6 formed from the 26 letters (without repetition) where either the first or last letter (possibly both) is not a vowel (a,e,i,o,u)? ANS: 1st way: 2*21*5*P(24, 4)?(one end cons., one vowel) + 21*20*P(24, 4) (both cons.) 2nd way: P(26,6) – P(5,2)P(24,4) (what cannot appear in 1st and last position); 3rd way: 2x21xP(25,5) {Consonant only in first position + Cons only in last} - P(21,2)P(24,4) {Cons in both (counted twice in first two terms)}4. How many 10-card hands are there chosen from a standard 52-card deck in which there are exactly two four-of-a-kinds; no pairs and no three-of-a-kinds? ANS: C(13,2) (pick which 2 of the 13 kinds have 4-of-a-kind)?C(4,4)2 (pick the 4 cards for each of these 2 kinds)?C(11,2)(pick which 2 kinds have 1-of-a-kind)?C(4,1)2 (pick the card for each of these 2 kinds)5. How many arrangements of HAPPINESS are there in which SE appear consecutively or SA appear consecutively but not both SE and SA are consecutive? ANS: Similar to problem #3- 2nd soln but now with both SA and SE are NOT allowed2x8!/2! (glue SA or SE (just one) and arrange)– 2x 7!/2 (glue both SA and SE). Outcomes with both SA and SE are counted twice in first term, but we do not want to count them at all. 6. How many arrangements of the letters in SIMULTANEOUS have ALL of the following 3 properties: (i) begins with an E (ii) at least two letters between each U, (iii) the consonants are in alphabetical order (iv) the vowels are not in alphabetical order.ANS: List 12 letters: A,E,I,O,U,U, L,M,N, S,S,T. First, ignore E- we will put it first after rest of arrangement is constructed; so 11 letters to arrange. Second, (iv) can be ignored because arrangements must start with E (not alphabetical order)- but still must arrange A,I,O. Third, find pattern of U’s and other 9 letters: 3 boxes before, between and after U’s with at least 2 letters in middle box: C(3 + (9-2)-1,(9-2)). Next pick 6 of 9 non-U positions for consonants- C(9,6) ways. Next arrange consonants- 1 way- and arrange other vowels- 3! ways.7. There are 8 Broadway musicals and they offer a special three-night package (Friday, Saturday, Sunday nights) where one can order one ticket that is good for three different musicals on successive nights (a sequence of three different musicals). A travel agent is going to order 30 of these tickets for a tour group of 30 people. How many ways are there to order a subset of 30 such tickets. ANS: There are 8x7x6 different tickets, i.e., sequences of 3 different musicals. There C(8x7x6,30) different subsets of 30 tickets. AMS 301 Second Test Spring 20201. (8 pt) Find the coefficient of x23 in (x3+x4+x5+x6) 4. = [ x3(1+ x+ x2+ x3)]4 = x12[(1 – x4)4/(1 – x)4] want coefficient of x23-12= x11 (1 – x4)4 = 1 – C(4,1)x4 + C(4,2)x8 + . . [1/(1-x)]4 = . . . C(r+4-1,r)xr a0b11 + a4b7 + a8b3 Coeff. of x11 = 1?C(11+4-1,11) - C(4,1)?C(7+4-1,7) + C(4,2)?C(3+4-1,3).2. (8 pts) Consider the problem of counting the ways to select 27 donuts from 5 types with at least three of each type.a) (2 pts) Model this problem as an integer-solution-of-equation-problem. e1 + . + e5=27, ei >=3b) (3 pts) Model this problem as a certain coefficient of a generating function. (x3+ . . . )5, coef of x27c) (3 pts) Solve this problem. C(5+(27-15)-1,(27-15))3. (9 pts) How many arrangements of length 12 formed by different letters (no repetition) chosen from the 26-letter alphabet are there that contain the five vowels (a,e,i,o,u)? Pick which 7 consonants are used and now arrange the 12 letters C(21,7)?12!; or Pick which positions for vowels and arrange the vowels, then pick consonants for the remaining 7 positions C(12,5)?5!?P(21,7) or C(12,5)?5!?C(21,7)x7!4. (8 pts) Find the probability that a 11-card hand (from a 52-card deck) has exactly 3 three-of-a-kinds (no 4-of-a-kinds and no pairs). {1/C(52,11)}[C(13,3) (pick which 3 of the 13 kinds have 3-of-a-kind)?C(4,3)3 (pick the 3 cards for each of these 3 kinds)?C(10,2)(pick which 2 kinds have 1-of-a-kind)?C(4,1)2 (pick the card for each of these 2 kinds)]5. (10 pts) How many ways are there to arrange 12 (distinct) faculty in a row so that Dr. Tucker is 4 positions away from President Bernstein (i.e., 3 people are inbetween Dr. Tucker and Dr. Bernstein), . . . T _ _ _ B . .. .? 2 (Tucker or Bernstein first)?8(places where the first one can go)x10!(arrange other 10 people in other 10 positions); can also do by 2-box model (one box before DrB/T, one inbetween whose size is fixed, and one after DrB/T).6. (15 pts) How many arrangements of 12 letters in ASSOCIATIONS have ALL of the following properties: (i) there are at least 2 letters inbetween each S, (ii) vowels are in alphabetical order, and (iii) the arrangement starts with an A. C(4 + (8-4)-1,(8-4))(distribute 8 non-S's (excluding A) into 4 boxes before, between and after the S's with at least 2 non-S's between each S ? C(8,5) (pick locations for other vowels) ?1(put vowels in alphabetical order) ?3!(arrange other consonants)7. (11 pts) How many ways are there to select 8 donuts from 6 types of donuts, if at most 3 donuts are chosen from the first three types combined; that is, at least 5 donuts chosen from the other three types. (Hint: break into cases) Break into 4 cases depending on how many donuts of first three types: sum k=0 to 3 of C(3+k-1,k)?C(3+(8-k)-1,(8-k) (ways to select k donuts from the first 3 types) ? (ways to select 8-k donuts from last 3 types)AMS 301.1A Prof. Tucker SECOND TEST Fall 2019 1. Find the coefficient of x29 in (x2+ x3+ x4+ x?+ x?)9. = [ x2(1+ x+ x2+ x3+ x4)]9 = x18[(1 – x5)9/(1 – x)9] want coefficient of x29-18= x11 (1 – x5)9 = 1 – C(9,1)x5 + C(9,2)x10 + . . [1/(1-x)]9 = . . . C(r+9-1,r)xr Ans: 1?C(11+9-1,11) - C(9,1) ?C(6+9-1,6) + C(9,2) ?C(1+9-1,1).2. Consider the problem of counting the ways to distribute 29 identical objects into 6 different boxes with at least three objects for each box.a) Model this problem as an integer-solution-of-equation-problem. . e1 + . + e6=29, ei >=3b) Model this problem as a certain coefficient of a generating function. (x3+ . . . )6, coef of x29c) Solve this problem. C(6+(29-18)-1,(29-18))3. How many 5-digit telephone numbers (5-digit decimal sequences) are there in which the digit 1 appears at most twice (maybe not at all)? Break into cases based on how many times 1 appears None once twice 95 + C(5,1)?94 + C(5,2)?934. Find the probability that a 12-card hand (from a 52-card deck) has exactly 3 two-of-a-kinds (no 4-of-a-kinds and no 3-of-a-kinds). (1/C(52,12)}[C(13,3)(pick which 3 of the 13 kinds have 2-of-a-kind)?C(4,2)^3(pick the 2 cards for each of these 3 kinds)?C(10,6)(pick which 6 kinds have 1-of-a-kind)?C(4,1)^6(pick the card for each of these 6 kinds)]5. How many sequences of length 10 are there with at least 3 vowels (a,e,i,o,u) and at least 6 consonants; repetition is allowed. Break into two cases: 3 vowels, 7 cons. and 4 vowels, 6 cons C(10,3) ? 53 ? 217 + C(10,4) ? 54 ? 216; Places for v’s seq of 3 v’s seq of 7 cons.6. How many arrangements of 15 letters in DIFFERENTIATION have ALL of the following properties: (i) there are at least 3 letters inbetween each I, (ii) the consonants are not in alphabetical order, and (iii) the arrangement starts with a consonant. First list the 15 letters: 7 vowels: I,I,I plus A,E,E,O 8 cons: DFFNNRTT C(4 + (11-6)-1,(11-6))(distribute 11 non-I’s (excluding leading cons.) into 4 boxes before, between and after the 3 I’s with at least 3 non-I’s between each I ? C(11,7) (pick locations for other 7 cons) ? (8!/2!2!2! – 1)(put cons. in non-alphabetical order) ? 4!/2!(arrange other vowels)7. How many subsets of four numbers from the set 2,3,4,5,6,7,8,9,10,11,12,13,14 are there in which the sum of the largest and smallest number in the subset is 15? Once largest and smallest number chosen, it remains to pick the 2 numbers inbetween 2 + 13 3 + 12 4 + 11 5 + 10 6 + 9 C(10,2) + C(8,2) + C(6,2) + C(4,2) + C(2,2) ................
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