Combinations, Permutations & Extended Binomial - IB Mathematics Analysis and Approaches

1. For what values of $x$ is the binomial expansion of $(9-3x)^{-1/2}$ valid?
   Answer: $|x| < 3$ The binomial expansion for $(1+u)^n$ where $n$ is not a positive integer is valid for $|u| < 1$. First, we must rewrite the given expression in this form. $(9-3x)^{-1/2} = [9(1-\frac{3x}{9})]^{-1/2} = 9^{-1/2}(1-\frac{x}{3})^{-1/2} = \frac{1}{3}(1-\frac{x}{3})^{-1/2}$. In this form, $u = -\frac{x}{3}$. The condition for validity is $|-\frac{x}{3}| < 1$, which simplifies to $\frac{|x|}{3} < 1$, or $|x| < 3$.
2. A team of 4 is to be chosen from 5 men and 6 women. How many distinct teams can be formed if there must be exactly two men on the team?
   Answer: 150 The condition is to choose exactly two men and consequently exactly two women to form a team of 4. Step 1: Choose 2 men from the 5 available. Since the order of selection does not matter, this is a combination: $\binom{5}{2} = \frac{5!}{2!3!} = 10$. Step 2: Choose 2 women from the 6 available. This is also a combination: $\binom{6}{2} = \frac{6!}{2!4!} = 15$. To find the total number of distinct teams, we multiply the number of ways for each step: $10 \times 15 = 150$.
3. Find the coefficient of $x^4$ in the expansion of $(1+2x)^5 (1-x)^{-2}$.
   Answer: 405 We need to find the terms up to $x^4$ in each expansion. For $(1+2x)^5$: $\binom{5}{0}(2x)^0 + \binom{5}{1}(2x)^1 + \binom{5}{2}(2x)^2 + \binom{5}{3}(2x)^3 + \binom{5}{4}(2x)^4 + \dots = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + \dots$ For $(1-x)^{-2}$: $1 + (-2)(-x) + \frac{(-2)(-3)}{2!}(-x)^2 + \frac{(-2)(-3)(-4)}{3!}(-x)^3 + \frac{(-2)(-3)(-4)(-5)}{4!}(-x)^4 + \dots = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots$ To find the coefficient of $x^4$ in the product, we multiply terms from each expansion whose powers of $x$ sum to 4: (Constant from 1st) × ($x^4$ from 2nd): $1 \times 5 = 5$ ($x$ from 1st) × ($x^3$ from 2nd): $10 \times 4 = 40$ ($x^2$ from 1st) × ($x^2$ from 2nd): $40 \times 3 = 120$ ($x^3$ from 1st) × ($x$ from 2nd): $80 \times 2 = 160$ ($x^4$ from 1st) × (Constant from 2nd): $80 \times 1 = 80$ The total coefficient is the sum: $5 + 40 + 120 + 160 + 80 = 405$.
4. Use the first three terms of the binomial expansion of $10(1-x)^{1/2}$ to find a rational approximation for $\sqrt{99}$.
   Answer: $\frac{79599}{8000}$ 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 $\sqrt{99}$, notice that $\sqrt{99} = \sqrt{100-1} = 10\sqrt{1-\frac{1}{100}} = 10(1-0.01)^{1/2}$. This matches the given form $10(1-x)^{1/2}$ with $x = 0.01$. Using the binomial expansion $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + ...$ with $u = -x = -0.01$ and $n = \frac{1}{2}$: $(1-x)^{1/2} = 1 + \frac{1}{2}(-x) + \frac{\frac{1}{2}(-\frac{1}{2})}{2}(-x)^2 + ...$ $= 1 - \frac{x}{2} - \frac{x^2}{8} + ...$ Substituting $x = 0.01$: $10(1-0.01)^{1/2} = 10(1 - 0.005 - 0.0000125) = 10(0.9949875) = 9.949875$ Converting to a fraction: $9.949875 = \frac{79599}{8000}$, which is answer D. Option A gives $\frac{199}{20} = 9.95$, which uses only the first two terms. Option B represents $10(1 - 0.005 + 0.0000125) = 9.950125$, incorrectly adding the third term. Option C gives $9.95125$, another computational error. Strategy tip: Always verify your binomial coefficient calculations carefully, especially the signs. The alternating pattern in $(1-x)^{1/2}$ means the third term subtracts, and small errors in these coefficients create the wrong answer choices.
5. How many ways can the letters of the word 'SUCCESS' be arranged such that the two 'C's are not together?
   Answer: 300 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 = $\frac{7!}{3!2!} = \frac{5040}{6 \times 2} = 420$. 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 = $\frac{6!}{3!} = \frac{720}{6} = 120$. 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 = $420 - 120 = 300$.
6. The first three terms in the expansion of $(1+ax)^n$ in ascending powers of $x$ are $1 + 6x + 24x^2$. Given that $n$ is not a positive integer, find the value of $n$.
   Answer: $-3$ 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 $(1+ax)^n = \sum_{k=0}^{\infty} \binom{n}{k}(ax)^k$ where the binomial coefficients are defined as $\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$. Let's use the given terms to find both $a$ and $n$. The expansion gives us: - Constant term: $\binom{n}{0}(ax)^0 = 1$ ✓ - Coefficient of $x$: $\binom{n}{1}(ax)^1 = na = 6$ - Coefficient of $x^2$: $\binom{n}{2}(ax)^2 = \frac{n(n-1)}{2}a^2 = 24$ From the second equation: $na = 6$, so $a = \frac{6}{n}$ Substituting into the third equation: $\frac{n(n-1)}{2} \cdot \left(\frac{6}{n}\right)^2 = 24$ Simplifying: $\frac{n(n-1)}{2} \cdot \frac{36}{n^2} = 24$ This gives us: $\frac{36(n-1)}{2n} = 24$, which simplifies to $18(n-1) = 24n$ Solving: $18n - 18 = 24n$, so $-18 = 6n$, therefore $n = -3$ Answer (D) $n = -3$ is correct. Answer (A) $n = -4$ would give different coefficients when you work through the algebra. Answer (B) $n = 4$ contradicts the given condition that n is not a positive integer. Answer (C) $n = 3$ 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.
7. Find the value of the sum $\sum_{k=0}^{10} \binom{10}{k} 3^k$.
   Answer: $4^{10}$ The question asks for the value of a sum that matches the form of the binomial theorem. The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. The given sum is $\sum_{k=0}^{10} \binom{10}{k} 3^k$. We can rewrite the term inside the summation as $\binom{10}{k} (1)^{10-k} (3)^k$. This perfectly matches the binomial expansion with $n=10$, $a=1$, and $b=3$. Therefore, the sum is equal to $(1+3)^{10} = 4^{10}$.
8. A group of 4 boys and 3 girls are to be seated in a single row. How many arrangements are possible if all 3 girls must sit together?
   Answer: 720 To solve this, we can treat the 3 girls who must sit together as a single block. Now, we are arranging 4 boys and this 1 block of girls. This is equivalent to arranging 5 distinct items. The number of ways to arrange these 5 items is $5! = 120$. Within the block of girls, the 3 girls can arrange themselves in $3! = 6$ ways. To get the total number of arrangements, we multiply the number of ways to arrange the blocks by the number of ways to arrange the individuals within the block: $5! \times 3! = 120 \times 6 = 720$.
9. How many distinct 5-digit numbers greater than 20000 can be formed using the digits 1, 1, 2, 3, 4?
   Answer: 36 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: $\frac{5!}{2!} = \frac{120}{2} = 60$. 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 $4! = 24$ such arrangements. The number of arrangements greater than 20000 is the total minus those starting with 1: $60 - 24 = 36$. Casework Method: Case 1: First digit is 2. The remaining digits are 1, 1, 3, 4. Arrangements = $\frac{4!}{2!} = 12$. Case 2: First digit is 3. The remaining digits are 1, 1, 2, 4. Arrangements = $\frac{4!}{2!} = 12$. Case 3: First digit is 4. The remaining digits are 1, 1, 2, 3. Arrangements = $\frac{4!}{2!} = 12$. Total = $12 + 12 + 12 = 36$.
10. 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?
    Answer: 2880 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 $5! = 120$ ways. - The 3 physics books can be arranged in $3! = 6$ ways. - The 2 chemistry books can be arranged in $2! = 2$ 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 = $2 \times 5! \times 3! \times 2! = 2 \times 120 \times 6 \times 2 = 2880$.
11. Find the coefficient of $x^3$ in the binomial expansion of $\frac{1}{\sqrt{4-x}}$.
    Answer: $\frac{5}{2048}$ First, rewrite the expression in the form $a(1+u)^n$: $\frac{1}{\sqrt{4-x}} = (4-x)^{-1/2} = 4^{-1/2}(1 - \frac{x}{4})^{-1/2} = \frac{1}{2}(1 - \frac{x}{4})^{-1/2}$. The general term in the expansion of $(1+u)^n$ is $\binom{n}{k}u^k = \frac{n(n-1)...(n-k+1)}{k!}u^k$. Here, $n = -1/2$ and we want the term with $x^3$, so we use $k=3$ and $u = -\frac{x}{4}$. The term in $u^3$ is $\frac{(-1/2)(-1/2-1)(-1/2-2)}{3!}u^3 = \frac{(-1/2)(-3/2)(-5/2)}{6}u^3 = \frac{-15/8}{6}u^3 = -\frac{15}{48}u^3 = -\frac{5}{16}u^3$. Now substitute $u = -\frac{x}{4}$: $-\frac{5}{16}(-\frac{x}{4})^3 = -\frac{5}{16}(-\frac{x^3}{64}) = \frac{5x^3}{1024}$. Finally, multiply by the initial factor of $\frac{1}{2}$: $\frac{1}{2} \times \frac{5x^3}{1024} = \frac{5x^3}{2048}$. The coefficient of $x^3$ is $\frac{5}{2048}$.
12. A committee of 3 men and 3 women is to be seated around a circular table. How many distinct seating arrangements are possible if the men and women must alternate?
    Answer: 12 First, seat one group, for example, the 3 men. The number of ways to arrange 3 distinct people in a circle is $(3-1)! = 2! = 2$. Once the men are seated, there are 3 specific empty seats between them. The 3 women must be seated in these 3 empty seats. Since the women are distinct individuals, they can be arranged in these 3 fixed seats in $3! = 6$ ways. The total number of arrangements is the product of these two steps: $2! \times 3! = 2 \times 6 = 12$.
13. In how many ways can a committee of 5 be formed from 10 people, if two particular people, A and B, refuse to serve on the committee together?
    Answer: 196 This problem can be solved using the complement method. First, find the total number of ways to form a committee of 5 from 10 people, without any restrictions. This is $\binom{10}{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252$. Next, find the number of ways to form a committee where A and B are together. If A and B are on the committee, we must choose them, and then select the remaining $5-2=3$ members from the remaining $10-2=8$ people. The number of ways to do this is $\binom{8}{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$. The number of committees where A and B are not together is the total number of committees minus the number where they are together: $252 - 56 = 196$.
14. How many distinct arrangements of the letters of the word LOGARITHM are there if no two vowels (O, A, I) can be adjacent?
    Answer: $6! \times P(7,3)$ 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 $6!$ 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 $P(7,3) = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5$, since order matters for the vowels. The total arrangements are $6! \times P(7,3)$, which is answer D. Answer A uses $\binom{7}{3}$, which only counts ways to choose positions but ignores that vowel order matters. Answer B incorrectly calculates $P(9,3)$, 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.
15. Find the coefficient of $x^2$ in the binomial expansion of $\sqrt[3]{8-x}$.
    Answer: $-\frac{1}{288}$ First, rewrite the expression in the form $a(1+u)^n$. $\sqrt[3]{8-x} = (8-x)^{1/3} = [8(1-\frac{x}{8})]^{1/3} = 8^{1/3}(1-\frac{x}{8})^{1/3} = 2(1-\frac{x}{8})^{1/3}$. Now, expand $(1-\frac{x}{8})^{1/3}$ using the binomial theorem for rational exponents, $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$, with $n=1/3$ and $u=-x/8$. The term with $x^2$ corresponds to $k=2$. The coefficient of $u^2$ is $\frac{n(n-1)}{2!} = \frac{(1/3)(1/3-1)}{2} = \frac{(1/3)(-2/3)}{2} = \frac{-2/9}{2} = -\frac{1}{9}$. The term is $-\frac{1}{9}u^2 = -\frac{1}{9}(-\frac{x}{8})^2 = -\frac{1}{9}(\frac{x^2}{64}) = -\frac{x^2}{576}$. Finally, we must multiply the entire expansion by the factor of 2 we extracted earlier. $2 \times ( -\frac{x^2}{576} ) = -\frac{2x^2}{576} = -\frac{x^2}{288}$. The coefficient of $x^2$ is $-\frac{1}{288}$.
16. In how many distinct ways can 3 identical red beads and 4 identical blue beads be arranged on a circular string?
    Answer: 5 This is a circular permutation problem with non-distinct items. The number of linear arrangements is $\binom{7}{3} = \frac{7!}{3!4!} = 35$. For circular arrangements where the number of items $n$ is prime (here, $n=7$), and not all items are identical, none of the arrangements will have rotational symmetry (other than the identity). Therefore, each distinct circular arrangement corresponds to exactly $n$ linear arrangements. We can find the number of distinct circular arrangements by dividing the number of linear arrangements by $n$. Number of ways = $\frac{1}{n} \binom{n}{k} = \frac{1}{7} \binom{7}{3} = \frac{35}{7} = 5$. This method is a simplification of Burnside's Lemma applicable when $n$ is prime and $k < n$.
17. The binomial expansion of $f(x) = (1-4x)^{-1/2}$ is valid for $|x|<a$, and the binomial expansion of $g(x) = (8+x)^{1/3}$ is valid for $|x|<b$. For which interval of $x$ is the expansion of the product $f(x)g(x)$ valid?
    Answer: $|x| < 1/4$ 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 $f(x) = (1-4x)^{-1/2}$, the expansion is valid when $|-4x| < 1$, which simplifies to $4|x| < 1$, or $|x| < 1/4$. So, $a = 1/4$. 2. For $g(x) = (8+x)^{1/3}$, we first factor out 8: $g(x) = 8^{1/3}(1 + \frac{x}{8})^{1/3} = 2(1 + \frac{x}{8})^{1/3}$. The expansion is valid when $|\frac{x}{8}| < 1$, which simplifies to $|x| < 8$. So, $b=8$. The product expansion $f(x)g(x)$ is valid only when $x$ is in the intersection of both intervals. We need $|x| < 1/4$ AND $|x| < 8$. The more restrictive of these two conditions is $|x| < 1/4$. Therefore, the product expansion is valid for $-1/4 < x < 1/4$.
18. A polygon has 44 diagonals. How many sides does it have?
    Answer: 11 Let the polygon have $n$ sides and $n$ vertices. The number of line segments that can be drawn between any two of its $n$ vertices is given by the combination formula $\binom{n}{2}$. These line segments consist of the sides of the polygon and its diagonals. Since there are $n$ sides, the number of diagonals is given by $\binom{n}{2} - n$. We are given that the number of diagonals is 44. So, $\binom{n}{2} - n = 44$. $\frac{n(n-1)}{2} - n = 44$. Multiply by 2: $n(n-1) - 2n = 88$. $n^2 - n - 2n = 88$. $n^2 - 3n - 88 = 0$. 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. $(n-11)(n+8) = 0$. The possible solutions are $n=11$ or $n=-8$. Since the number of sides must be a positive integer, $n=11$.
19. 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?
    Answer: 265 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 = $\binom{5}{3} \times \binom{6}{4} = 10 \times 15 = 150$. Case 2: 4 Mathematicians and 3 Physicists. Number of ways = $\binom{5}{4} \times \binom{6}{3} = 5 \times 20 = 100$. Case 3: 5 Mathematicians and 2 Physicists. Number of ways = $\binom{5}{5} \times \binom{6}{2} = 1 \times 15 = 15$. The total number of possible committees is the sum of the ways for each case: $150 + 100 + 15 = 265$.
20. 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?
    Answer: 805 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 $\binom{12}{6} = 924$. 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): $\binom{7}{6} = 7$. 2. Selections with NO blue balls (i.e., from 5 red + 3 green = 8 balls): $\binom{8}{6} = \binom{8}{2} = 28$. 3. Selections with NO green balls (i.e., from 5 red + 4 blue = 9 balls): $\binom{9}{6} = \binom{9}{3} = 84$. The number of selections missing two colors (e.g., no red and no blue) is $\binom{3}{6} = 0$, so there are no overlaps between these cases. The total number of 'bad' selections is the sum: $7 + 28 + 84 = 119$. The number of valid selections is Total - Bad = $924 - 119 = 805$.

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