IB Mathematics: Analysis and Approaches Quiz: Combinatorics Basics
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Combinatorics BasicsQuestion 1 of 20

What is the value of the sum ∑r=010(10r)(−1)r\sum_{r=0}^{10} \binom{10}{r} (-1)^r?

-1
0
1
1024
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Combinatorics Basics

Practice Combinatorics Basics 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 Combinatorics Basics, 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

What is the value of the sum ∑r=010(10r)(−1)r\sum_{r=0}^{10} \binom{10}{r} (-1)^r?

  1. -1
  2. 0 (correct answer)
  3. 1
  4. 1024
Explanation: This sum is a direct application of the binomial theorem. The binomial theorem states that (a+b)n=∑r=0n(nr)an−rbr(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r. Let's choose specific values for aa, bb, and nn to match the given sum. The sum is ∑r=010(10r)(−1)r\sum_{r=0}^{10} \binom{10}{r} (-1)^r. We can rewrite this as ∑r=010(10r)(1)10−r(−1)r\sum_{r=0}^{10} \binom{10}{r} (1)^{10-r} (-1)^r. This perfectly matches the expansion of (a+b)n(a+b)^n with n=10n=10, a=1a=1, and b=−1b=-1. So, the sum is equal to (1+(−1))10=(1−1)10=010=0(1 + (-1))^{10} = (1-1)^{10} = 0^{10} = 0. In general, for any integer n≥1n \ge 1, the alternating sum of binomial coefficients ∑r=0n(nr)(−1)r=0\sum_{r=0}^{n} \binom{n}{r} (-1)^r = 0. Distractor D (1024) is the value of ∑r=010(10r)=210\sum_{r=0}^{10} \binom{10}{r} = 2^{10}.

Question 2

In the expansion of (1+x)n(1+x)^n, the coefficient of x2x^2 is three times the coefficient of xx. Find the value of nn.

  1. 4
  2. 5
  3. 6
  4. 7 (correct answer)
Explanation: The expansion of (1+x)n(1+x)^n begins 1+(n1)x+(n2)x2+…1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dots. The coefficient of xx is c1=(n1)=nc_1 = \binom{n}{1} = n. The coefficient of x2x^2 is c2=(n2)=n(n−1)2c_2 = \binom{n}{2} = \frac{n(n-1)}{2}. The problem states that c2=3c1c_2 = 3c_1. Substituting the expressions for the coefficients, we get: n(n−1)2=3n\frac{n(n-1)}{2} = 3n. Since nn must be at least 2 for an x2x^2 term to exist, n≠0n \neq 0, so we can divide both sides by nn: n−12=3\frac{n-1}{2} = 3. Multiply both sides by 2: n−1=6n-1 = 6. Therefore, n=7n=7. Distractor A (n=4) would arise if one forgot the denominator 2 in the formula for (n2)\binom{n}{2}, leading to the incorrect equation n(n−1)=3nn(n-1)=3n, which simplifies to n−1=3n-1=3.

Question 3

What is the value of the sum ∑r=010(10r)(−1)r\sum_{r=0}^{10} \binom{10}{r} (-1)^r?

  1. -1
  2. 0 (correct answer)
  3. 1
  4. 1024
Explanation: This sum is a direct application of the binomial theorem. The binomial theorem states that (a+b)n=∑r=0n(nr)an−rbr(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r. Let's choose specific values for aa, bb, and nn to match the given sum. The sum is ∑r=010(10r)(−1)r\sum_{r=0}^{10} \binom{10}{r} (-1)^r. We can rewrite this as ∑r=010(10r)(1)10−r(−1)r\sum_{r=0}^{10} \binom{10}{r} (1)^{10-r} (-1)^r. This perfectly matches the expansion of (a+b)n(a+b)^n with n=10n=10, a=1a=1, and b=−1b=-1. So, the sum is equal to (1+(−1))10=(1−1)10=010=0(1 + (-1))^{10} = (1-1)^{10} = 0^{10} = 0. In general, for any integer n≥1n \ge 1, the alternating sum of binomial coefficients ∑r=0n(nr)(−1)r=0\sum_{r=0}^{n} \binom{n}{r} (-1)^r = 0. Distractor D (1024) is the value of ∑r=010(10r)=210\sum_{r=0}^{10} \binom{10}{r} = 2^{10}.

Question 4

The coefficient of x5x^5 in the expansion of (2x+a)8(2x+a)^8 is 14336. Find the value of aa.

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 8
Explanation: The general term in the expansion of (2x+a)8(2x+a)^8 is Tr+1=(8r)(2x)ra8−rT_{r+1} = \binom{8}{r}(2x)^r a^{8-r}. We want the term with x5x^5, so we set r=5r=5. The term is T6=(85)(2x)5a8−5=(85)25x5a3T_6 = \binom{8}{5}(2x)^5 a^{8-5} = \binom{8}{5} 2^5 x^5 a^3. The coefficient of x5x^5 is (85)25a3\binom{8}{5} 2^5 a^3. First, calculate the binomial coefficient: (85)=(83)=8×7×63×2×1=56\binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. Next, calculate the power of 2: 25=322^5 = 32. So the coefficient is 56×32×a3=1792a356 \times 32 \times a^3 = 1792 a^3. We are given that this coefficient is 14336. So, 1792a3=143361792 a^3 = 14336. Divide to solve for a3a^3: a3=143361792=8a^3 = \frac{14336}{1792} = 8. Taking the cube root, we find a=2a=2. Distractor C arises from a calculation error, for instance, using 242^4 instead of 252^5, which would lead to 896a3=14336896a^3=14336, so a3=16a^3=16, which might be mistaken as a=4a=4. Distractor D arises from taking a=a3a=a^3.

Question 5

In the expansion of (4+x)29(4+x)^{29}, the coefficients of xkx^k and xk+1x^{k+1} are equal. Find the value of kk.

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 23
Explanation: The general term in the expansion of (4+x)29(4+x)^{29} is Tr+1=(29r)429−rxrT_{r+1} = \binom{29}{r} 4^{29-r} x^r. The coefficient of xkx^k is obtained when r=kr=k, which is (29k)429−k\binom{29}{k} 4^{29-k}. The coefficient of xk+1x^{k+1} is obtained when r=k+1r=k+1, which is (29k+1)429−(k+1)=(29k+1)428−k\binom{29}{k+1} 4^{29-(k+1)} = \binom{29}{k+1} 4^{28-k}. We are given that these coefficients are equal: (29k)429−k=(29k+1)428−k\binom{29}{k} 4^{29-k} = \binom{29}{k+1} 4^{28-k}. Divide both sides by 428−k4^{28-k}: (29k)⋅4=(29k+1)\binom{29}{k} \cdot 4 = \binom{29}{k+1}. Now expand the binomial coefficients using the formula (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}: 29!k!(29−k)!⋅4=29!(k+1)!(29−(k+1))!\frac{29!}{k!(29-k)!} \cdot 4 = \frac{29!}{(k+1)!(29-(k+1))!}. 4k!(29−k)(28−k)!=1(k+1)k!(28−k)!\frac{4}{k!(29-k)(28-k)!} = \frac{1}{(k+1)k!(28-k)!}. Cancel common terms k!k! and (28−k)!(28-k)!: 429−k=1k+1\frac{4}{29-k} = \frac{1}{k+1}. Cross-multiply: 4(k+1)=29−k4(k+1) = 29-k. 4k+4=29−k4k+4 = 29-k. 5k=255k = 25. k=5k=5. Distractor D (23) comes from incorrectly setting up the ratio as 129−k=4k+1\frac{1}{29-k} = \frac{4}{k+1}.

Question 6

The third term in the expansion of (1+ax)n(1+ax)^n is 36x236x^2 and the fourth term is 84x384x^3. Find the value of nn.

  1. 6
  2. 7
  3. 8
  4. 9 (correct answer)
Explanation: The general term in the expansion of (1+ax)n(1+ax)^n is Tr+1=(nr)(ax)r=(nr)arxrT_{r+1} = \binom{n}{r}(ax)^r = \binom{n}{r}a^r x^r. The third term corresponds to r=2r=2: T3=(n2)a2x2T_3 = \binom{n}{2}a^2x^2. The coefficient is (n2)a2=36\binom{n}{2}a^2 = 36. The fourth term corresponds to r=3r=3: T4=(n3)a3x3T_4 = \binom{n}{3}a^3x^3. The coefficient is (n3)a3=84\binom{n}{3}a^3 = 84. Let's form a ratio of the two coefficients: (n3)a3(n2)a2=8436\frac{\binom{n}{3}a^3}{\binom{n}{2}a^2} = \frac{84}{36}. This simplifies to n!3!(n−3)!n!2!(n−2)!⋅a=73\frac{\frac{n!}{3!(n-3)!}}{\frac{n!}{2!(n-2)!}} \cdot a = \frac{7}{3}. n!3!(n−3)!⋅2!(n−2)!n!⋅a=73\frac{n!}{3!(n-3)!} \cdot \frac{2!(n-2)!}{n!} \cdot a = \frac{7}{3}. 2!(n−2)(n−3)!3⋅2!(n−3)!⋅a=73\frac{2!(n-2)(n-3)!}{3 \cdot 2!(n-3)!} \cdot a = \frac{7}{3}. n−23⋅a=73\frac{n-2}{3} \cdot a = \frac{7}{3}, which simplifies to a(n−2)=7a(n-2) = 7. From the third term's coefficient: n(n−1)2a2=36\frac{n(n-1)}{2}a^2 = 36. From a=7n−2a = \frac{7}{n-2}, we substitute: n(n−1)2(7n−2)2=36\frac{n(n-1)}{2} \left(\frac{7}{n-2}\right)^2 = 36. n(n−1)249(n−2)2=36\frac{n(n-1)}{2} \frac{49}{(n-2)^2} = 36. This looks complicated. Let's test the integer options. If n=9n=9, then a(9−2)=7  ⟹  7a=7  ⟹  a=1a(9-2)=7 \implies 7a=7 \implies a=1. Check with the third term: (92)(1)2=9×82=36\binom{9}{2}(1)^2 = \frac{9 \times 8}{2} = 36. This works. Check with the fourth term: (93)(1)3=9×8×73×2×1=84\binom{9}{3}(1)^3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84. This also works. Thus, n=9n=9.

Question 7

Which of the following expressions is equivalent to (kr)+2(kr+1)+(kr+2)\binom{k}{r} + 2\binom{k}{r+1} + \binom{k}{r+2}?

  1. (k+1r+1)\binom{k+1}{r+1}
  2. (k+2r+1)\binom{k+2}{r+1}
  3. (k+2r+2)\binom{k+2}{r+2} (correct answer)
  4. (kr+1)+(k+1r+2)\binom{k}{r+1} + \binom{k+1}{r+2}
Explanation: This problem uses Pascal's identity, which states that (nk)+(nk+1)=(n+1k+1)\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}. We can rewrite the given expression by splitting the middle term: (kr)+2(kr+1)+(kr+2)=[(kr)+(kr+1)]+[(kr+1)+(kr+2)]\binom{k}{r} + 2\binom{k}{r+1} + \binom{k}{r+2} = \left[ \binom{k}{r} + \binom{k}{r+1} \right] + \left[ \binom{k}{r+1} + \binom{k}{r+2} \right]. Now, apply Pascal's identity to each pair of terms in the brackets:
  1. (kr)+(kr+1)=(k+1r+1)\binom{k}{r} + \binom{k}{r+1} = \binom{k+1}{r+1}
  2. (kr+1)+(kr+2)=(k+1r+2)\binom{k}{r+1} + \binom{k}{r+2} = \binom{k+1}{r+2} Substitute these back into the expression: (k+1r+1)+(k+1r+2)\binom{k+1}{r+1} + \binom{k+1}{r+2}. We can apply Pascal's identity one more time to this new expression (with n=k+1n=k+1 and k=r+1k=r+1): (k+1r+1)+(k+1r+2)=((k+1)+1(r+1)+1)=(k+2r+2)\binom{k+1}{r+1} + \binom{k+1}{r+2} = \binom{(k+1)+1}{(r+1)+1} = \binom{k+2}{r+2}. Therefore, the original expression is equivalent to (k+2r+2)\binom{k+2}{r+2}.

Question 8

In the expansion of (1+x)n(1+x)^n, the coefficient of x2x^2 is three times the coefficient of xx. Find the value of nn.

  1. 4
  2. 5
  3. 6
  4. 7 (correct answer)
Explanation: The expansion of (1+x)n(1+x)^n begins 1+(n1)x+(n2)x2+…1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dots. The coefficient of xx is c1=(n1)=nc_1 = \binom{n}{1} = n. The coefficient of x2x^2 is c2=(n2)=n(n−1)2c_2 = \binom{n}{2} = \frac{n(n-1)}{2}. The problem states that c2=3c1c_2 = 3c_1. Substituting the expressions for the coefficients, we get: n(n−1)2=3n\frac{n(n-1)}{2} = 3n. Since nn must be at least 2 for an x2x^2 term to exist, n≠0n \neq 0, so we can divide both sides by nn: n−12=3\frac{n-1}{2} = 3. Multiply both sides by 2: n−1=6n-1 = 6. Therefore, n=7n=7. Distractor A (n=4) would arise if one forgot the denominator 2 in the formula for (n2)\binom{n}{2}, leading to the incorrect equation n(n−1)=3nn(n-1)=3n, which simplifies to n−1=3n-1=3.

Question 9

Given that (n2)=105\binom{n}{2} = 105, what is the value of (n−13)\binom{n-1}{3}?

  1. 182
  2. 286
  3. 364 (correct answer)
  4. 455
Explanation: First, we need to find the value of nn from the equation (n2)=105\binom{n}{2} = 105. The formula for (n2)\binom{n}{2} is n(n−1)2\frac{n(n-1)}{2}. So, n(n−1)2=105\frac{n(n-1)}{2} = 105. n(n−1)=210n(n-1) = 210. We are looking for two consecutive integers whose product is 210. We can estimate: 210\sqrt{210} is between 196=14\sqrt{196}=14 and 225=15\sqrt{225}=15. Let's try n=15n=15. Then n(n−1)=15(14)=210n(n-1) = 15(14) = 210. This is correct, so n=15n=15. Now we need to calculate (n−13)\binom{n-1}{3} with n=15n=15. This is (15−13)=(143)\binom{15-1}{3} = \binom{14}{3}. (143)=14×13×123×2×1\binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1}. (143)=14×13×2=364\binom{14}{3} = 14 \times 13 \times 2 = 364. A common mistake is to calculate (153)\binom{15}{3}, which is 455455 (Distractor D).

Question 10

The coefficient of x8x^8 in the expansion of (x2+k)7(x^2+k)^7 is 280. Find the value of kk.

  1. 2 (correct answer)
  2. 4
  3. 5
  4. 7
Explanation: The general term in the expansion of (x2+k)7(x^2+k)^7 is Tr+1=(7r)(x2)7−rkrT_{r+1} = \binom{7}{r} (x^2)^{7-r} k^r. Simplifying the term: Tr+1=(7r)x14−2rkrT_{r+1} = \binom{7}{r} x^{14-2r} k^r. We want the term with x8x^8, so we set the exponent of xx to 8. 14−2r=814-2r = 8 6=2r6 = 2r r=3r=3. The coefficient of this term is obtained by substituting r=3r=3 into the coefficient part of the general term: (73)k3\binom{7}{3}k^3. Calculate the binomial coefficient: (73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35. The coefficient is 35k335k^3. We are given that this coefficient is 280. 35k3=28035k^3 = 280. k3=28035=8k^3 = \frac{280}{35} = 8. Taking the cube root, we find k=2k=2.

Question 11

Given that (nk−1)=36\binom{n}{k-1} = 36, (nk)=84\binom{n}{k} = 84, and (nk+1)=126\binom{n}{k+1} = 126 for some integers nn and kk. Find the value of (nn−k)\binom{n}{n-k}.

  1. 36
  2. 84 (correct answer)
  3. 90
  4. 126
Explanation: The question asks for the value of (nn−k)\binom{n}{n-k}. A key property of binomial coefficients is the symmetry identity: (nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}. Applying this identity with r=kr=k, we get (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k}. The problem statement gives the value of (nk)\binom{n}{k} as 84. Therefore, (nn−k)=84\binom{n}{n-k} = 84. The other values, (nk−1)=36\binom{n}{k-1} = 36 and (nk+1)=126\binom{n}{k+1} = 126, are extra information designed to distract the test-taker.

Question 12

Find the term independent of xx in the expansion of (2x2−1x)9\left( 2x^2 - \frac{1}{x} \right)^9.

  1. -5376
  2. 84
  3. 672 (correct answer)
  4. 5376
Explanation: The general term of the expansion (a+b)n(a+b)^n is given by Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r. Here, a=2x2a = 2x^2, b=−1xb = -\frac{1}{x}, and n=9n=9. The general term is Tr+1=(9r)(2x2)9−r(−1x)rT_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(-\frac{1}{x}\right)^r. Simplifying the term: Tr+1=(9r)29−r(x2)9−r(−1)r(x−1)r=(9r)29−r(−1)rx18−2rx−r=(9r)29−r(−1)rx18−3rT_{r+1} = \binom{9}{r} 2^{9-r} (x^2)^{9-r} (-1)^r (x^{-1})^r = \binom{9}{r} 2^{9-r} (-1)^r x^{18-2r} x^{-r} = \binom{9}{r} 2^{9-r} (-1)^r x^{18-3r}. The term independent of xx is the one where the exponent of xx is 0. Set 18−3r=018-3r = 0, which gives 3r=183r=18, so r=6r=6. Now substitute r=6r=6 back into the expression for the term: Term = (96)29−6(−1)6x18−3(6)\binom{9}{6} 2^{9-6} (-1)^6 x^{18-3(6)} = (93)23(1)x0\binom{9}{3} 2^3 (1) x^0. Calculate the value: (93)=9×8×73×2×1=84\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84. The term is 84×23=84×8=67284 \times 2^3 = 84 \times 8 = 672. Distractor A (-5376) could arise from miscalculating the power of x, for example finding r=3, which gives (93)26(−1)3=84×64×(−1)\binom{9}{3}2^6(-1)^3 = 84 \times 64 \times (-1). Distractor B (84) arises from forgetting the 29−r2^{9-r} factor. Distractor D could arise from a sign error or miscalculation.

Question 13

Which of the following expressions is equivalent to (kr)+2(kr+1)+(kr+2)\binom{k}{r} + 2\binom{k}{r+1} + \binom{k}{r+2}?

  1. (k+1r+1)\binom{k+1}{r+1}
  2. (k+2r+1)\binom{k+2}{r+1}
  3. (k+2r+2)\binom{k+2}{r+2} (correct answer)
  4. (kr+1)+(k+1r+2)\binom{k}{r+1} + \binom{k+1}{r+2}
Explanation: This problem uses Pascal's identity, which states that (nk)+(nk+1)=(n+1k+1)\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}. We can rewrite the given expression by splitting the middle term: (kr)+2(kr+1)+(kr+2)=[(kr)+(kr+1)]+[(kr+1)+(kr+2)]\binom{k}{r} + 2\binom{k}{r+1} + \binom{k}{r+2} = \left[ \binom{k}{r} + \binom{k}{r+1} \right] + \left[ \binom{k}{r+1} + \binom{k}{r+2} \right]. Now, apply Pascal's identity to each pair of terms in the brackets:
  1. (kr)+(kr+1)=(k+1r+1)\binom{k}{r} + \binom{k}{r+1} = \binom{k+1}{r+1}
  2. (kr+1)+(kr+2)=(k+1r+2)\binom{k}{r+1} + \binom{k}{r+2} = \binom{k+1}{r+2} Substitute these back into the expression: (k+1r+1)+(k+1r+2)\binom{k+1}{r+1} + \binom{k+1}{r+2}. We can apply Pascal's identity one more time to this new expression (with n=k+1n=k+1 and k=r+1k=r+1): (k+1r+1)+(k+1r+2)=((k+1)+1(r+1)+1)=(k+2r+2)\binom{k+1}{r+1} + \binom{k+1}{r+2} = \binom{(k+1)+1}{(r+1)+1} = \binom{k+2}{r+2}. Therefore, the original expression is equivalent to (k+2r+2)\binom{k+2}{r+2}.

Question 14

Consider the expansion of (x2+1)m(x^2+1)^m. The sum of the coefficients of the terms in the expansion is 1024. What is the degree of the resulting polynomial?

  1. 10
  2. 20 (correct answer)
  3. 512
  4. 1024
Explanation: The sum of the coefficients in any polynomial expansion P(x)P(x) can be found by evaluating P(1)P(1). For the expansion of (x2+1)m(x^2+1)^m, the sum of the coefficients is (12+1)m=(1+1)m=2m(1^2+1)^m = (1+1)^m = 2^m. We are given that this sum is 1024. So, 2m=10242^m = 1024. Since 1024=2101024 = 2^{10}, we can conclude that m=10m=10. The polynomial is (x2+1)10(x^2+1)^{10}. The degree of a polynomial is the highest power of the variable after expansion. The highest power of xx will come from the first term in the binomial expansion, which is (1010)(x2)10(1)0=(x2)10=x20\binom{10}{10}(x^2)^{10}(1)^0 = (x^2)^{10} = x^{20}. Therefore, the degree of the polynomial is 20. A common error is to choose 10, which is the value of mm, not the degree of the polynomial.

Question 15

The coefficient of the term a3b7a^3b^7 in the expansion of (a+b)10(a+b)^{10} is CC. What is the coefficient of the term x7y3x^7y^3 in the expansion of (x+y)10(x+y)^{10}?

  1. CC (correct answer)
  2. 1C\frac{1}{C}
  3. 37C\frac{3}{7}C
  4. C2C^2
Explanation: The coefficient of the term akbn−ka^k b^{n-k} in the expansion of (a+b)n(a+b)^n is given by (nk)\binom{n}{k}. Or, equivalently, the coefficient of akbja^k b^j where k+j=nk+j=n is (nk)=(nj)\binom{n}{k} = \binom{n}{j}. For the term a3b7a^3b^7 in (a+b)10(a+b)^{10}, n=10n=10, the power of aa is 3, and the power of bb is 7. The coefficient is (103)\binom{10}{3} or (107)\binom{10}{7}. We are given that this value is CC. So, C=(103)=(107)C = \binom{10}{3} = \binom{10}{7}. For the term x7y3x^7y^3 in the expansion of (x+y)10(x+y)^{10}, n=10n=10, the power of xx is 7, and the power of yy is 3. The coefficient is (107)\binom{10}{7} or (103)\binom{10}{3}. Since (107)=(103)\binom{10}{7} = \binom{10}{3}, the coefficient of x7y3x^7y^3 is also CC. This question tests the symmetry property of binomial coefficients: (nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}.

Question 16

What is the coefficient of xx in the expansion of (x3−2x)7\left( x^3 - \frac{2}{x} \right)^7?

  1. -672 (correct answer)
  2. -336
  3. 336
  4. 672
Explanation: The general term in the expansion of (a+b)n(a+b)^n is Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r. For (x3−2x)7\left( x^3 - \frac{2}{x} \right)^7, we have a=x3a=x^3, b=−2xb=-\frac{2}{x}, and n=7n=7. The general term is Tr+1=(7r)(x3)7−r(−2x)rT_{r+1} = \binom{7}{r} (x^3)^{7-r} \left( -\frac{2}{x} \right)^r. Let's simplify the expression to find the power of xx: Tr+1=(7r)x3(7−r)(−2)rx−r=(7r)(−2)rx21−3r−r=(7r)(−2)rx21−4rT_{r+1} = \binom{7}{r} x^{3(7-r)} (-2)^r x^{-r} = \binom{7}{r} (-2)^r x^{21-3r-r} = \binom{7}{r} (-2)^r x^{21-4r}. We want the coefficient of xx, which means the exponent of xx should be 1. Set 21−4r=121-4r = 1. 20=4r20 = 4r. r=5r=5. Now substitute r=5r=5 into the coefficient part of the general term: (75)(−2)5\binom{7}{5}(-2)^5. Calculate the parts: (75)=(72)=7×62=21\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2} = 21 and (−2)5=−32(-2)^5 = -32. The coefficient is 21×(−32)=−67221 \times (-32) = -672. Distractor D (672) is a common error from forgetting the negative sign on (−2)5(-2)^5.

Question 17

Given that (n2)=105\binom{n}{2} = 105, what is the value of (n−13)\binom{n-1}{3}?

  1. 182
  2. 286
  3. 364 (correct answer)
  4. 455
Explanation: First, we need to find the value of nn from the equation (n2)=105\binom{n}{2} = 105. The formula for (n2)\binom{n}{2} is n(n−1)2\frac{n(n-1)}{2}. So, n(n−1)2=105\frac{n(n-1)}{2} = 105. n(n−1)=210n(n-1) = 210. We are looking for two consecutive integers whose product is 210. We can estimate: 210\sqrt{210} is between 196=14\sqrt{196}=14 and 225=15\sqrt{225}=15. Let's try n=15n=15. Then n(n−1)=15(14)=210n(n-1) = 15(14) = 210. This is correct, so n=15n=15. Now we need to calculate (n−13)\binom{n-1}{3} with n=15n=15. This is (15−13)=(143)\binom{15-1}{3} = \binom{14}{3}. (143)=14×13×123×2×1\binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1}. (143)=14×13×2=364\binom{14}{3} = 14 \times 13 \times 2 = 364. A common mistake is to calculate (153)\binom{15}{3}, which is 455455 (Distractor D).

Question 18

In the expansion of (1+x)n(1+x)^n, the coefficients of the 4th and 6th terms are equal. What is the value of nn?

  1. 7
  2. 8 (correct answer)
  3. 9
  4. 10
Explanation: In the expansion of (1+x)n(1+x)^n, the general term is Tr+1=(nr)xrT_{r+1} = \binom{n}{r}x^r. The coefficient is (nr)\binom{n}{r}. The 4th term corresponds to r=3r=3, so its coefficient is (n3)\binom{n}{3}. The 6th term corresponds to r=5r=5, so its coefficient is (n5)\binom{n}{5}. We are given that these coefficients are equal: (n3)=(n5)\binom{n}{3} = \binom{n}{5}. We use the identity (na)=(nb)\binom{n}{a} = \binom{n}{b} which implies either a=ba=b (not possible here as 3≠53 \neq 5) or a+b=na+b=n. Therefore, n=3+5=8n = 3+5 = 8. Distractor A (n=7) might result from an off-by-one error, comparing (n4)\binom{n}{4} and (n6)\binom{n}{6} and then making another error. For example, 4+6=104+6=10 and then dividing by something.

Question 19

In the expansion of (1+x)n(1+x)^n, the coefficients of the 4th and 6th terms are equal. What is the value of nn?

  1. 7
  2. 8 (correct answer)
  3. 9
  4. 10
Explanation: In the expansion of (1+x)n(1+x)^n, the general term is Tr+1=(nr)xrT_{r+1} = \binom{n}{r}x^r. The coefficient is (nr)\binom{n}{r}. The 4th term corresponds to r=3r=3, so its coefficient is (n3)\binom{n}{3}. The 6th term corresponds to r=5r=5, so its coefficient is (n5)\binom{n}{5}. We are given that these coefficients are equal: (n3)=(n5)\binom{n}{3} = \binom{n}{5}. We use the identity (na)=(nb)\binom{n}{a} = \binom{n}{b} which implies either a=ba=b (not possible here as 3≠53 \neq 5) or a+b=na+b=n. Therefore, n=3+5=8n = 3+5 = 8. Distractor A (n=7) might result from an off-by-one error, comparing (n4)\binom{n}{4} and (n6)\binom{n}{6} and then making another error. For example, 4+6=104+6=10 and then dividing by something.

Question 20

In the expansion of (4+x)29(4+x)^{29}, the coefficients of xkx^k and xk+1x^{k+1} are equal. Find the value of kk.

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 23
Explanation: The general term in the expansion of (4+x)29(4+x)^{29} is Tr+1=(29r)429−rxrT_{r+1} = \binom{29}{r} 4^{29-r} x^r. The coefficient of xkx^k is obtained when r=kr=k, which is (29k)429−k\binom{29}{k} 4^{29-k}. The coefficient of xk+1x^{k+1} is obtained when r=k+1r=k+1, which is (29k+1)429−(k+1)=(29k+1)428−k\binom{29}{k+1} 4^{29-(k+1)} = \binom{29}{k+1} 4^{28-k}. We are given that these coefficients are equal: (29k)429−k=(29k+1)428−k\binom{29}{k} 4^{29-k} = \binom{29}{k+1} 4^{28-k}. Divide both sides by 428−k4^{28-k}: (29k)⋅4=(29k+1)\binom{29}{k} \cdot 4 = \binom{29}{k+1}. Now expand the binomial coefficients using the formula (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}: 29!k!(29−k)!⋅4=29!(k+1)!(29−(k+1))!\frac{29!}{k!(29-k)!} \cdot 4 = \frac{29!}{(k+1)!(29-(k+1))!}. 4k!(29−k)(28−k)!=1(k+1)k!(28−k)!\frac{4}{k!(29-k)(28-k)!} = \frac{1}{(k+1)k!(28-k)!}. Cancel common terms k!k! and (28−k)!(28-k)!: 429−k=1k+1\frac{4}{29-k} = \frac{1}{k+1}. Cross-multiply: 4(k+1)=29−k4(k+1) = 29-k. 4k+4=29−k4k+4 = 29-k. 5k=255k = 25. k=5k=5. Distractor D (23) comes from incorrectly setting up the ratio as 129−k=4k+1\frac{1}{29-k} = \frac{4}{k+1}.