IB Mathematics: Analysis and Approaches Quiz: Combinations Permutations And Extended Binomial
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Combinations Permutations And Extended BinomialQuestion 1 of 20

A committee of 7 people is to be selected from a group of 5 mathematicians and 6 physicists. How many different committees can be formed if the committee must contain at least 3 mathematicians and at least 2 physicists?

165
265
330
480
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Combinations Permutations And Extended Binomial

Practice Combinations Permutations And Extended Binomial in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Combinations Permutations And Extended Binomial, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

A committee of 7 people is to be selected from a group of 5 mathematicians and 6 physicists. How many different committees can be formed if the committee must contain at least 3 mathematicians and at least 2 physicists?

  1. 165
  2. 265 (correct answer)
  3. 330
  4. 480
Explanation: The committee size is 7. The conditions are at least 3 mathematicians (M) and at least 2 physicists (P). We list the possible compositions of the committee that satisfy the conditions: Case 1: 3 Mathematicians and 4 Physicists. Number of ways = (53)×(64)=10×15=150\binom{5}{3} \times \binom{6}{4} = 10 \times 15 = 150. Case 2: 4 Mathematicians and 3 Physicists. Number of ways = (54)×(63)=5×20=100\binom{5}{4} \times \binom{6}{3} = 5 \times 20 = 100. Case 3: 5 Mathematicians and 2 Physicists. Number of ways = (55)×(62)=1×15=15\binom{5}{5} \times \binom{6}{2} = 1 \times 15 = 15. The total number of possible committees is the sum of the ways for each case: 150+100+15=265150 + 100 + 15 = 265.

Question 2

Five distinct mathematics books, three distinct physics books, and two distinct chemistry books are to be arranged on a shelf. How many arrangements are possible if the books of each subject must stay together and the physics books must be in the middle block?

  1. 128
  2. 1440
  3. 2880 (correct answer)
  4. 8640
Explanation: First, treat the books of each subject as a single block: a Math block (M), a Physics block (P), and a Chemistry block (C). The problem states that the Physics block must be in the middle. This leaves two possible arrangements for the blocks: M P C or C P M. So there are 2 ways to arrange the blocks. Next, we find the number of ways to arrange the distinct books within each block:
  • The 5 math books can be arranged in 5!=1205! = 120 ways.
  • The 3 physics books can be arranged in 3!=63! = 6 ways.
  • The 2 chemistry books can be arranged in 2!=22! = 2 ways. The total number of arrangements is the product of the number of ways to arrange the blocks and the number of ways to arrange the books within each block: Total ways = 2×5!×3!×2!=2×120×6×2=28802 \times 5! \times 3! \times 2! = 2 \times 120 \times 6 \times 2 = 2880.

Question 3

A committee of 7 people is to be selected from a group of 5 mathematicians and 6 physicists. How many different committees can be formed if the committee must contain at least 3 mathematicians and at least 2 physicists?

  1. 165
  2. 265 (correct answer)
  3. 330
  4. 480
Explanation: The committee size is 7. The conditions are at least 3 mathematicians (M) and at least 2 physicists (P). We list the possible compositions of the committee that satisfy the conditions: Case 1: 3 Mathematicians and 4 Physicists. Number of ways = (53)×(64)=10×15=150\binom{5}{3} \times \binom{6}{4} = 10 \times 15 = 150. Case 2: 4 Mathematicians and 3 Physicists. Number of ways = (54)×(63)=5×20=100\binom{5}{4} \times \binom{6}{3} = 5 \times 20 = 100. Case 3: 5 Mathematicians and 2 Physicists. Number of ways = (55)×(62)=1×15=15\binom{5}{5} \times \binom{6}{2} = 1 \times 15 = 15. The total number of possible committees is the sum of the ways for each case: 150+100+15=265150 + 100 + 15 = 265.

Question 4

The binomial expansion of f(x)=(1−4x)−1/2f(x) = (1-4x)^{-1/2} is valid for ∣x∣<a|x|<a, and the binomial expansion of g(x)=(8+x)1/3g(x) = (8+x)^{1/3} is valid for ∣x∣<b|x|<b. For which interval of xx is the expansion of the product f(x)g(x)f(x)g(x) valid?

  1. ∣x∣<1/4|x| < 1/4 (correct answer)
  2. ∣x∣<2|x| < 2
  3. ∣x∣<8|x| < 8
  4. ∣x∣<32|x| < 32
Explanation: For the expansion of a product of two series to be valid, both individual series expansions must be valid. We need to find the interval of convergence for each function.
  1. For f(x)=(1−4x)−1/2f(x) = (1-4x)^{-1/2}, the expansion is valid when ∣−4x∣<1|-4x| < 1, which simplifies to 4∣x∣<14|x| < 1, or ∣x∣<1/4|x| < 1/4. So, a=1/4a = 1/4.
  2. For g(x)=(8+x)1/3g(x) = (8+x)^{1/3}, we first factor out 8: g(x)=81/3(1+x8)1/3=2(1+x8)1/3g(x) = 8^{1/3}(1 + \frac{x}{8})^{1/3} = 2(1 + \frac{x}{8})^{1/3}. The expansion is valid when ∣x8∣<1|\frac{x}{8}| < 1, which simplifies to ∣x∣<8|x| < 8. So, b=8b=8. The product expansion f(x)g(x)f(x)g(x) is valid only when xx is in the intersection of both intervals. We need ∣x∣<1/4|x| < 1/4 AND ∣x∣<8|x| < 8. The more restrictive of these two conditions is ∣x∣<1/4|x| < 1/4. Therefore, the product expansion is valid for −1/4<x<1/4-1/4 < x < 1/4.

Question 5

How many distinct 5-digit numbers greater than 20000 can be formed using the digits 1, 1, 2, 3, 4?

  1. 12
  2. 24
  3. 36 (correct answer)
  4. 60
Explanation: For the 5-digit number to be greater than 20000, the first digit must be 2, 3, or 4. We can analyze this by cases, or by using the complement method. Complement Method: First find the total number of distinct 5-digit numbers. We are arranging 5 digits where '1' is repeated twice: 5!2!=1202=60\frac{5!}{2!} = \frac{120}{2} = 60. Now find the number of arrangements that are NOT greater than 20000. This means the number must start with 1. If the first digit is 1, we need to arrange the remaining four digits: 1, 2, 3, 4. These are all distinct, so there are 4!=244! = 24 such arrangements. The number of arrangements greater than 20000 is the total minus those starting with 1: 60−24=3660 - 24 = 36. Casework Method: Case 1: First digit is 2. The remaining digits are 1, 1, 3, 4. Arrangements = 4!2!=12\frac{4!}{2!} = 12. Case 2: First digit is 3. The remaining digits are 1, 1, 2, 4. Arrangements = 4!2!=12\frac{4!}{2!} = 12. Case 3: First digit is 4. The remaining digits are 1, 1, 2, 3. Arrangements = 4!2!=12\frac{4!}{2!} = 12. Total = 12+12+12=3612 + 12 + 12 = 36.

Question 6

A polygon has 44 diagonals. How many sides does it have?

  1. 8
  2. 9
  3. 10
  4. 11 (correct answer)
Explanation: Let the polygon have nn sides and nn vertices. The number of line segments that can be drawn between any two of its nn vertices is given by the combination formula (n2)\binom{n}{2}. These line segments consist of the sides of the polygon and its diagonals. Since there are nn sides, the number of diagonals is given by (n2)−n\binom{n}{2} - n. We are given that the number of diagonals is 44. So, (n2)−n=44\binom{n}{2} - n = 44. n(n−1)2−n=44\frac{n(n-1)}{2} - n = 44. Multiply by 2: n(n−1)−2n=88n(n-1) - 2n = 88. n2−n−2n=88n^2 - n - 2n = 88. n2−3n−88=0n^2 - 3n - 88 = 0. We can solve this quadratic equation by factoring. We look for two numbers that multiply to -88 and add to -3. These are -11 and 8. (n−11)(n+8)=0(n-11)(n+8) = 0. The possible solutions are n=11n=11 or n=−8n=-8. Since the number of sides must be a positive integer, n=11n=11.

Question 7

The binomial expansion of f(x)=(1−4x)−1/2f(x) = (1-4x)^{-1/2} is valid for ∣x∣<a|x|<a, and the binomial expansion of g(x)=(8+x)1/3g(x) = (8+x)^{1/3} is valid for ∣x∣<b|x|<b. For which interval of xx is the expansion of the product f(x)g(x)f(x)g(x) valid?

  1. ∣x∣<1/4|x| < 1/4 (correct answer)
  2. ∣x∣<2|x| < 2
  3. ∣x∣<8|x| < 8
  4. ∣x∣<32|x| < 32
Explanation: For the expansion of a product of two series to be valid, both individual series expansions must be valid. We need to find the interval of convergence for each function.
  1. For f(x)=(1−4x)−1/2f(x) = (1-4x)^{-1/2}, the expansion is valid when ∣−4x∣<1|-4x| < 1, which simplifies to 4∣x∣<14|x| < 1, or ∣x∣<1/4|x| < 1/4. So, a=1/4a = 1/4.
  2. For g(x)=(8+x)1/3g(x) = (8+x)^{1/3}, we first factor out 8: g(x)=81/3(1+x8)1/3=2(1+x8)1/3g(x) = 8^{1/3}(1 + \frac{x}{8})^{1/3} = 2(1 + \frac{x}{8})^{1/3}. The expansion is valid when ∣x8∣<1|\frac{x}{8}| < 1, which simplifies to ∣x∣<8|x| < 8. So, b=8b=8. The product expansion f(x)g(x)f(x)g(x) is valid only when xx is in the intersection of both intervals. We need ∣x∣<1/4|x| < 1/4 AND ∣x∣<8|x| < 8. The more restrictive of these two conditions is ∣x∣<1/4|x| < 1/4. Therefore, the product expansion is valid for −1/4<x<1/4-1/4 < x < 1/4.

Question 8

Five distinct mathematics books, three distinct physics books, and two distinct chemistry books are to be arranged on a shelf. How many arrangements are possible if the books of each subject must stay together and the physics books must be in the middle block?

  1. 128
  2. 1440
  3. 2880 (correct answer)
  4. 8640
Explanation: First, treat the books of each subject as a single block: a Math block (M), a Physics block (P), and a Chemistry block (C). The problem states that the Physics block must be in the middle. This leaves two possible arrangements for the blocks: M P C or C P M. So there are 2 ways to arrange the blocks. Next, we find the number of ways to arrange the distinct books within each block:
  • The 5 math books can be arranged in 5!=1205! = 120 ways.
  • The 3 physics books can be arranged in 3!=63! = 6 ways.
  • The 2 chemistry books can be arranged in 2!=22! = 2 ways. The total number of arrangements is the product of the number of ways to arrange the blocks and the number of ways to arrange the books within each block: Total ways = 2×5!×3!×2!=2×120×6×2=28802 \times 5! \times 3! \times 2! = 2 \times 120 \times 6 \times 2 = 2880.

Question 9

A bag contains 5 red, 4 blue, and 3 green balls. In how many ways can 6 balls be selected if there must be at least one ball of each color?

  1. 119
  2. 805 (correct answer)
  3. 840
  4. 924
Explanation: This problem is best solved using the principle of inclusion-exclusion or the complement method. Total number of ways to select 6 balls from the 12 available is (126)=924\binom{12}{6} = 924. The condition is 'at least one of each color'. The complement is 'missing at least one color'. Let's calculate the number of selections that violate the condition:
  1. Selections with NO red balls (i.e., from 4 blue + 3 green = 7 balls): (76)=7\binom{7}{6} = 7.
  2. Selections with NO blue balls (i.e., from 5 red + 3 green = 8 balls): (86)=(82)=28\binom{8}{6} = \binom{8}{2} = 28.
  3. Selections with NO green balls (i.e., from 5 red + 4 blue = 9 balls): (96)=(93)=84\binom{9}{6} = \binom{9}{3} = 84. The number of selections missing two colors (e.g., no red and no blue) is (36)=0\binom{3}{6} = 0, so there are no overlaps between these cases. The total number of 'bad' selections is the sum: 7+28+84=1197 + 28 + 84 = 119. The number of valid selections is Total - Bad = 924−119=805924 - 119 = 805.

Question 10

Find the value of the sum ∑k=010(10k)3k\sum_{k=0}^{10} \binom{10}{k} 3^k.

  1. 3103^{10}
  2. 3113^{11}
  3. 4104^{10} (correct answer)
  4. 4114^{11}
Explanation: The question asks for the value of a sum that matches the form of the binomial theorem. The binomial theorem states that (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. The given sum is ∑k=010(10k)3k\sum_{k=0}^{10} \binom{10}{k} 3^k. We can rewrite the term inside the summation as (10k)(1)10−k(3)k\binom{10}{k} (1)^{10-k} (3)^k. This perfectly matches the binomial expansion with n=10n=10, a=1a=1, and b=3b=3. Therefore, the sum is equal to (1+3)10=410(1+3)^{10} = 4^{10}.

Question 11

The first three terms in the expansion of (1+ax)n(1+ax)^n in ascending powers of xx are 1+6x+24x21 + 6x + 24x^2. Given that nn is not a positive integer, find the value of nn.

  1. −4-4
  2. 44
  3. 33
  4. −3-3 (correct answer)
Explanation: When you encounter a binomial expansion where n might not be a positive integer, you're dealing with the generalized binomial theorem. This extends beyond the familiar (1+ax)n=∑k=0∞(nk)(ax)k(1+ax)^n = \sum_{k=0}^{\infty} \binom{n}{k}(ax)^k where the binomial coefficients are defined as (nk)=n(n−1)(n−2)...(n−k+1)k!\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}. Let's use the given terms to find both aa and nn. The expansion gives us:
  • Constant term: (n0)(ax)0=1\binom{n}{0}(ax)^0 = 1 ✓
  • Coefficient of xx: (n1)(ax)1=na=6\binom{n}{1}(ax)^1 = na = 6
  • Coefficient of x2x^2: (n2)(ax)2=n(n−1)2a2=24\binom{n}{2}(ax)^2 = \frac{n(n-1)}{2}a^2 = 24
From the second equation: na=6na = 6, so a=6na = \frac{6}{n} Substituting into the third equation: n(n−1)2⋅(6n)2=24\frac{n(n-1)}{2} \cdot \left(\frac{6}{n}\right)^2 = 24 Simplifying: n(n−1)2⋅36n2=24\frac{n(n-1)}{2} \cdot \frac{36}{n^2} = 24 This gives us: 36(n−1)2n=24\frac{36(n-1)}{2n} = 24, which simplifies to 18(n−1)=24n18(n-1) = 24n Solving: 18n−18=24n18n - 18 = 24n, so −18=6n-18 = 6n, therefore n=−3n = -3 Answer (D) n=−3n = -3 is correct. Answer (A) n=−4n = -4 would give different coefficients when you work through the algebra. Answer (B) n=4n = 4 contradicts the given condition that n is not a positive integer. Answer (C) n=3n = 3 also violates this condition. Study tip: When working with generalized binomial expansions, always check that your final answer satisfies all given conditions, including any restrictions on the parameters.

Question 12

Find the coefficient of x4x^4 in the expansion of (1+2x)5(1−x)−2(1+2x)^5 (1-x)^{-2}.

  1. 80
  2. 325
  3. 405 (correct answer)
  4. 485
Explanation: We need to find the terms up to x4x^4 in each expansion. For (1+2x)5(1+2x)^5: (50)(2x)0+(51)(2x)1+(52)(2x)2+(53)(2x)3+(54)(2x)4+⋯=1+10x+40x2+80x3+80x4+…\binom{5}{0}(2x)^0 + \binom{5}{1}(2x)^1 + \binom{5}{2}(2x)^2 + \binom{5}{3}(2x)^3 + \binom{5}{4}(2x)^4 + \dots = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + \dots For (1−x)−2(1-x)^{-2}: 1+(−2)(−x)+(−2)(−3)2!(−x)2+(−2)(−3)(−4)3!(−x)3+(−2)(−3)(−4)(−5)4!(−x)4+⋯=1+2x+3x2+4x3+5x4+…1 + (-2)(-x) + \frac{(-2)(-3)}{2!}(-x)^2 + \frac{(-2)(-3)(-4)}{3!}(-x)^3 + \frac{(-2)(-3)(-4)(-5)}{4!}(-x)^4 + \dots = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots To find the coefficient of x4x^4 in the product, we multiply terms from each expansion whose powers of xx sum to 4: (Constant from 1st) × (x4x^4 from 2nd): 1×5=51 \times 5 = 5 (xx from 1st) × (x3x^3 from 2nd): 10×4=4010 \times 4 = 40 (x2x^2 from 1st) × (x2x^2 from 2nd): 40×3=12040 \times 3 = 120 (x3x^3 from 1st) × (xx from 2nd): 80×2=16080 \times 2 = 160 (x4x^4 from 1st) × (Constant from 2nd): 80×1=8080 \times 1 = 80 The total coefficient is the sum: 5+40+120+160+80=4055 + 40 + 120 + 160 + 80 = 405.

Question 13

How many distinct 5-digit numbers greater than 20000 can be formed using the digits 1, 1, 2, 3, 4?

  1. 12
  2. 24
  3. 36 (correct answer)
  4. 60
Explanation: For the 5-digit number to be greater than 20000, the first digit must be 2, 3, or 4. We can analyze this by cases, or by using the complement method. Complement Method: First find the total number of distinct 5-digit numbers. We are arranging 5 digits where '1' is repeated twice: 5!2!=1202=60\frac{5!}{2!} = \frac{120}{2} = 60. Now find the number of arrangements that are NOT greater than 20000. This means the number must start with 1. If the first digit is 1, we need to arrange the remaining four digits: 1, 2, 3, 4. These are all distinct, so there are 4!=244! = 24 such arrangements. The number of arrangements greater than 20000 is the total minus those starting with 1: 60−24=3660 - 24 = 36. Casework Method: Case 1: First digit is 2. The remaining digits are 1, 1, 3, 4. Arrangements = 4!2!=12\frac{4!}{2!} = 12. Case 2: First digit is 3. The remaining digits are 1, 1, 2, 4. Arrangements = 4!2!=12\frac{4!}{2!} = 12. Case 3: First digit is 4. The remaining digits are 1, 1, 2, 3. Arrangements = 4!2!=12\frac{4!}{2!} = 12. Total = 12+12+12=3612 + 12 + 12 = 36.

Question 14

How many ways can the letters of the word 'SUCCESS' be arranged such that the two 'C's are not together?

  1. 120
  2. 300 (correct answer)
  3. 420
  4. 600
Explanation: This problem can be solved using the complement method. First, find the total number of arrangements of the letters in 'SUCCESS'. There are 7 letters in total, with 3 'S's and 2 'C's. Total arrangements = 7!3!2!=50406×2=420\frac{7!}{3!2!} = \frac{5040}{6 \times 2} = 420. Next, find the number of arrangements where the two 'C's ARE together. Treat 'CC' as a single block. We are now arranging the items {S, S, S, U, E, (CC)}. This is 6 items, with 3 'S's. Arrangements with 'CC' together = 6!3!=7206=120\frac{6!}{3!} = \frac{720}{6} = 120. The number of arrangements where the 'C's are not together is the total number of arrangements minus the number where they are together. Ways = 420−120=300420 - 120 = 300.

Question 15

Find the coefficient of x2x^2 in the binomial expansion of 8−x3\sqrt[3]{8-x}.

  1. −1288-\frac{1}{288} (correct answer)
  2. −1576-\frac{1}{576}
  3. −29-\frac{2}{9}
  4. 1288\frac{1}{288}
Explanation: First, rewrite the expression in the form a(1+u)na(1+u)^n. 8−x3=(8−x)1/3=[8(1−x8)]1/3=81/3(1−x8)1/3=2(1−x8)1/3\sqrt[3]{8-x} = (8-x)^{1/3} = [8(1-\frac{x}{8})]^{1/3} = 8^{1/3}(1-\frac{x}{8})^{1/3} = 2(1-\frac{x}{8})^{1/3}. Now, expand (1−x8)1/3(1-\frac{x}{8})^{1/3} using the binomial theorem for rational exponents, (1+u)n=1+nu+n(n−1)2!u2+…(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots, with n=1/3n=1/3 and u=−x/8u=-x/8. The term with x2x^2 corresponds to k=2k=2. The coefficient of u2u^2 is n(n−1)2!=(1/3)(1/3−1)2=(1/3)(−2/3)2=−2/92=−19\frac{n(n-1)}{2!} = \frac{(1/3)(1/3-1)}{2} = \frac{(1/3)(-2/3)}{2} = \frac{-2/9}{2} = -\frac{1}{9}. The term is −19u2=−19(−x8)2=−19(x264)=−x2576-\frac{1}{9}u^2 = -\frac{1}{9}(-\frac{x}{8})^2 = -\frac{1}{9}(\frac{x^2}{64}) = -\frac{x^2}{576}. Finally, we must multiply the entire expansion by the factor of 2 we extracted earlier. 2×(−x2576)=−2x2576=−x22882 \times ( -\frac{x^2}{576} ) = -\frac{2x^2}{576} = -\frac{x^2}{288}. The coefficient of x2x^2 is −1288-\frac{1}{288}.

Question 16

Find the value of the sum ∑k=010(10k)3k\sum_{k=0}^{10} \binom{10}{k} 3^k.

  1. 3103^{10}
  2. 3113^{11}
  3. 4104^{10} (correct answer)
  4. 4114^{11}
Explanation: The question asks for the value of a sum that matches the form of the binomial theorem. The binomial theorem states that (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. The given sum is ∑k=010(10k)3k\sum_{k=0}^{10} \binom{10}{k} 3^k. We can rewrite the term inside the summation as (10k)(1)10−k(3)k\binom{10}{k} (1)^{10-k} (3)^k. This perfectly matches the binomial expansion with n=10n=10, a=1a=1, and b=3b=3. Therefore, the sum is equal to (1+3)10=410(1+3)^{10} = 4^{10}.

Question 17

How many distinct arrangements of the letters of the word LOGARITHM are there if no two vowels (O, A, I) can be adjacent?

  1. 6!×(73)6! \times \binom{7}{3}
  2. P(9,3)×6!P(9,3) \times 6!
  3. 9!−3!×7!9! - 3! \times 7!
  4. 6!×P(7,3)6! \times P(7,3) (correct answer)
Explanation: This question tests your ability to handle restricted permutations, specifically the "no adjacency" constraint for certain elements. When you see vowels that can't be adjacent, think of it as a two-step process: arrange the consonants first, then place vowels in the gaps. LOGARITHM has 9 letters total: 6 consonants (L, G, R, T, H, M) and 3 vowels (O, A, I). To ensure no vowels are adjacent, first arrange the 6 consonants in a line, which can be done in 6!6! ways. These 6 consonants create 7 possible positions for vowels: before the first consonant, between each pair of consonants, and after the last consonant. Next, you need to select 3 of these 7 positions for your vowels and arrange the vowels in those positions. This is P(7,3)=7!(7−3)!=7!4!=7×6×5P(7,3) = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5, since order matters for the vowels. The total arrangements are 6!×P(7,3)6! \times P(7,3), which is answer D. Answer A uses (73)\binom{7}{3}, which only counts ways to choose positions but ignores that vowel order matters. Answer B incorrectly calculates P(9,3)P(9,3), treating this as selecting 3 letters from 9 rather than placing vowels in consonant gaps. Answer C attempts a subtraction approach (total minus restricted cases) but miscalculates the restricted arrangements. Strategy tip: For "non-adjacent" problems, always use the gap method: arrange the unrestricted elements first, then place restricted elements in the created gaps. Remember to use permutations, not combinations, when order matters.

Question 18

The first three terms in the expansion of (1+ax)n(1+ax)^n in ascending powers of xx are 1+6x+24x21 + 6x + 24x^2. Given that nn is not a positive integer, find the value of nn.

  1. −4-4
  2. 44
  3. 33
  4. −3-3 (correct answer)
Explanation: When you encounter a binomial expansion where n might not be a positive integer, you're dealing with the generalized binomial theorem. This extends beyond the familiar (1+ax)n=∑k=0∞(nk)(ax)k(1+ax)^n = \sum_{k=0}^{\infty} \binom{n}{k}(ax)^k where the binomial coefficients are defined as (nk)=n(n−1)(n−2)...(n−k+1)k!\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}. Let's use the given terms to find both aa and nn. The expansion gives us:
  • Constant term: (n0)(ax)0=1\binom{n}{0}(ax)^0 = 1 ✓
  • Coefficient of xx: (n1)(ax)1=na=6\binom{n}{1}(ax)^1 = na = 6
  • Coefficient of x2x^2: (n2)(ax)2=n(n−1)2a2=24\binom{n}{2}(ax)^2 = \frac{n(n-1)}{2}a^2 = 24
From the second equation: na=6na = 6, so a=6na = \frac{6}{n} Substituting into the third equation: n(n−1)2⋅(6n)2=24\frac{n(n-1)}{2} \cdot \left(\frac{6}{n}\right)^2 = 24 Simplifying: n(n−1)2⋅36n2=24\frac{n(n-1)}{2} \cdot \frac{36}{n^2} = 24 This gives us: 36(n−1)2n=24\frac{36(n-1)}{2n} = 24, which simplifies to 18(n−1)=24n18(n-1) = 24n Solving: 18n−18=24n18n - 18 = 24n, so −18=6n-18 = 6n, therefore n=−3n = -3 Answer (D) n=−3n = -3 is correct. Answer (A) n=−4n = -4 would give different coefficients when you work through the algebra. Answer (B) n=4n = 4 contradicts the given condition that n is not a positive integer. Answer (C) n=3n = 3 also violates this condition. Study tip: When working with generalized binomial expansions, always check that your final answer satisfies all given conditions, including any restrictions on the parameters.

Question 19

Use the first three terms of the binomial expansion of 10(1−x)1/210(1-x)^{1/2} to find a rational approximation for 99\sqrt{99}.

  1. 19920\frac{199}{20}
  2. 799998000\frac{79999}{8000}
  3. 796018000\frac{79601}{8000}
  4. 795998000\frac{79599}{8000} (correct answer)
Explanation: When you encounter a problem asking for rational approximations using binomial expansions, you're dealing with the binomial theorem for non-integer exponents. The key insight is recognizing how to manipulate the expression to match a standard form and connect it to the value you're approximating. To find 99\sqrt{99}, notice that 99=100−1=101−1100=10(1−0.01)1/2\sqrt{99} = \sqrt{100-1} = 10\sqrt{1-\frac{1}{100}} = 10(1-0.01)^{1/2}. This matches the given form 10(1−x)1/210(1-x)^{1/2} with x=0.01x = 0.01. Using the binomial expansion (1+u)n=1+nu+n(n−1)2!u2+...(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + ... with u=−x=−0.01u = -x = -0.01 and n=12n = \frac{1}{2}: (1−x)1/2=1+12(−x)+12(−12)2(−x)2+...(1-x)^{1/2} = 1 + \frac{1}{2}(-x) + \frac{\frac{1}{2}(-\frac{1}{2})}{2}(-x)^2 + ... =1−x2−x28+...= 1 - \frac{x}{2} - \frac{x^2}{8} + ... Substituting x=0.01x = 0.01: 10(1−0.01)1/2=10(1−0.005−0.0000125)=10(0.9949875)=9.94987510(1-0.01)^{1/2} = 10(1 - 0.005 - 0.0000125) = 10(0.9949875) = 9.949875 Converting to a fraction: 9.949875=7959980009.949875 = \frac{79599}{8000}, which is answer D. Option A gives 19920=9.95\frac{199}{20} = 9.95, which uses only the first two terms. Option B represents 10(1−0.005+0.0000125)=9.95012510(1 - 0.005 + 0.0000125) = 9.950125, incorrectly adding the third term. Option C gives 9.951259.95125, another computational error. Strategy tip: Always verify your binomial coefficient calculations carefully, especially the signs. The alternating pattern in (1−x)1/2(1-x)^{1/2} means the third term subtracts, and small errors in these coefficients create the wrong answer choices.

Question 20

Find the coefficient of x4x^4 in the expansion of (1+2x)5(1−x)−2(1+2x)^5 (1-x)^{-2}.

  1. 80
  2. 325
  3. 405 (correct answer)
  4. 485
Explanation: We need to find the terms up to x4x^4 in each expansion. For (1+2x)5(1+2x)^5: (50)(2x)0+(51)(2x)1+(52)(2x)2+(53)(2x)3+(54)(2x)4+⋯=1+10x+40x2+80x3+80x4+…\binom{5}{0}(2x)^0 + \binom{5}{1}(2x)^1 + \binom{5}{2}(2x)^2 + \binom{5}{3}(2x)^3 + \binom{5}{4}(2x)^4 + \dots = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + \dots For (1−x)−2(1-x)^{-2}: 1+(−2)(−x)+(−2)(−3)2!(−x)2+(−2)(−3)(−4)3!(−x)3+(−2)(−3)(−4)(−5)4!(−x)4+⋯=1+2x+3x2+4x3+5x4+…1 + (-2)(-x) + \frac{(-2)(-3)}{2!}(-x)^2 + \frac{(-2)(-3)(-4)}{3!}(-x)^3 + \frac{(-2)(-3)(-4)(-5)}{4!}(-x)^4 + \dots = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots To find the coefficient of x4x^4 in the product, we multiply terms from each expansion whose powers of xx sum to 4: (Constant from 1st) × (x4x^4 from 2nd): 1×5=51 \times 5 = 5 (xx from 1st) × (x3x^3 from 2nd): 10×4=4010 \times 4 = 40 (x2x^2 from 1st) × (x2x^2 from 2nd): 40×3=12040 \times 3 = 120 (x3x^3 from 1st) × (xx from 2nd): 80×2=16080 \times 2 = 160 (x4x^4 from 1st) × (Constant from 2nd): 80×1=8080 \times 1 = 80 The total coefficient is the sum: 5+40+120+160+80=4055 + 40 + 120 + 160 + 80 = 405.