1.4 Binomial Theorems and Combinatorics

Q1

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	N17/5/MATHL/HP1/ENG/TZ0/XX/4
Question Text	Find the coefficient of $x^9$ in the expansion of $\left ( x^{2}+\frac{3}{x} \right )^6$
Total Mark	4
Correct Answer	18
Explanation	n/a
Mark Scheme	Step 1: Find the general form Each term is of the form Step 2: Find the value of the variable ${}^{6}C_r \times \left ( x^{2} \right )^{5}\left ( \frac{3}{x} \right )^{6-5}=6\times x^{9}$ , where $k$ is the coefficient of $x^8$ Take the exponent, $2r - (6-r) = 9$ , so $r = 5$ $kx^{9} = {}^{6}C_5\left(x^{2}\right)^{5}\left(\frac{3}{x}\right)^{6-5}= 6\times3\times x^{8}$ So $k = 18$

Q2

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	M17/5/MATHL/HP1/ENG/TZ2/XX/1
Question Text	Find the term independent of $x$ in the binomial expansion of $\left ( 3x^{2}+\frac{1}{43} \right )^{10}$ .
Total Mark	5
Correct Answer	1890
Explanation	n/a
Mark Scheme	Use binomial expansion, Pascal’s triangle, or the binomial coefficient to find the constant term in which $x$ in canceled out. The term independent of $x$ is $\left ( {}^{10}C_4 \right )\left ( 3x^{2} \right )^{6}\left ( \frac{1}{3x^{3}} \right )^{4} = 1890$ (or equivalent)

Q3

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	M16-TZ2-P1-6(HL)
Question Text	Consider the expansion of $(1 + x)^n$ in ascending powers of x, where $n\geq3$ . (a) Given that the fourth term of the expansion is expressed as $\frac{n\left ( n+a \right )\left ( n+b \right )}{c}x^{3}$ , find the value of $a + b + c$ .
Total Mark	2
Correct Answer	3
Explanation	n/a
Mark Scheme	When we expand the expression, the fourth term is equal to $\frac{n\left ( n-1 \right )\left ( n-2 \right )}{6}x^{3}$ Hence, $a=-1, b=-2, c=6$ Therefore, the sum of the three equal terms is equal to 3.
Question Text	(b) The coefficients of the second, third and fourth terms of the expansion are consecutive terms of an arithmetic squence. (i) Given that $n^3-9n^2+kn=0$ , find the value of $k$ . Total Mark: 3 Correct Answer: 14 Explanation: n/a Mark Scheme: Tip 1: Consider the given information. In this case, we are given that the 2nd, 3rd and 4th term of the expansion forms an arithmetic sequence. Using this information, we can form an equation using $n$ . $d = u_{3}-u_{2}=u_{4}-u_{3}$ $\frac{n(n-1)}{2}-n=\frac{n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}$ If we attempt to remove the denominators and expand the equation, we get $6n+n^3-3n^2+2n=6n^2-6n$ (or equivalent) $n^3-9n^2+14n=0$ Hence, the value of $k$ is equal to 14. (ii) Hence, find the value of $n$ . Total Mark: 1 Correct Answer: 7 Explanation: n/a Mark Scheme: If we solve the equation from the previous question, we can derive the value of $n$ . $n(n - 2)(n - 7)$ $= 0$ or $(n - 2)(n - 7) = 0$ Since we are given the condition that $n≥3$ , $n=7$

Q4

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	N15-TZ0-P1-3(HL)
Question Text	(a) Simplify the expansion of $(4+x)^5$ in ascending powers of $x$ and find the coefficients of $x^3$ .
Total Mark	1
Correct Answer	160
Explanation	n/a
Mark Scheme	Tip 1: Expand $(4+x)^5$ following the binomial theorem. $\left ( 4+x \right )^{5}=4^{5}+5\cdot 4^{4}x+10\cdot 4^{3}x^{2}+10\cdot 4^{2}x^{3}+5\cdot 4^{1}x^{4}+x^{5}$ Tip 2: Find the coefficient of the term that the question requires. The coefficient of $x^3$ is equal to $10\cdot 4^{2}=160$ .
Question Text	(b) Hence find the exact value of $(4.1)^5$ .
Total Mark	3
Correct Answer	1158.56201
Explanation	n/a
Mark Scheme	Use the binomial expansion from (a). If we let $x = 0.1$ in the binomial expansion, $4.1⁵ = 1024 + 128 + 6.4 + 0.16 + 0.002 + 0.00001 = 1158.56201$

Q5

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	M15-TZ2-P1-2(HL)
Question Text	Expand $(5 - x)⁴$ in ascending powers of $x$ and find the sum of all the term’s coefficients.
Total Mark	4
Correct Answer	156
Explanation	n/a
Mark Scheme	Expand $(5 - x)⁴$ using the binomial theorem and simplify. $\left ( 5-x \right )^{4}=5^{4}+4\cdot 5^{3}(-x)+6\cdot 5^{2}(-x)^{2}+4\cdot 5(-x)^{3}+(-x)^{4}$ $= 625 - 600x + 150x^2-20x^3+x^4$ Then, find the sum of all the term’s coefficients. $625-600+150-20+1=156$ Hence, the answer is 156

Q6

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	N14/5/MATHL/HP1/ENG/TZ0/XX/10
Question Text	A set of positive integers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is used to form a pack of ten cards. Each card displays one positive integer without repetition from this set. Grace wishes to select four cards at random from this pack of ten cards. (a) Find the number of selections Grace could make if the largest integer drawn among the four cards is either a 7, 8 or 9.
Total Mark	2
Correct Answer	111
Explanation	n/a
Mark Scheme	Use combinations to find the number of ways in which 7, 8, and 9 is the largest integer among the four cards and add them together. ${}^{6}C_3 + {}^{7}C_3 + {}^{8}C_3 = 20 + 35 + 56 = 111$
Question Text	(b) Find the number of selections Grace could make if at least two of the four integers drawn are even.
Total Mark	4
Correct Answer	155
Explanation	n/a
Mark Scheme	METHOD 1: To find the number of selections with at least two even integers, we can subtract the number of selections that do not satisfy the given condition from the total number possible selections. Total number of possible selections: ${}^{10}C_4= 210$ 4 odd: ${}^{5}C_4 = 5$ 3 odd and 1 even: ${}^{5}C_3 \times {}^{5}C_1 = 50$ Hence, the number of selections with at least two even integers is equal to $210 - 5 - 50 = 155$ METHOD 2 To have at least two even integers among the four integers, the following combinations are possible: 2 odd and 2 even: ${}^{5}C_2 \times {}^{5}C_2 = 100$ 1 odd and 3 even: ${}^{5}C_1 \times x {}^{5}C_3 = 50$ 0 odd and 4 even: ${}^{5}C_0 \times {}^{5}C_4 = 5$ Hence, the number of selections with at least two even integers is equal to $100 + 50 + 5 = 155$

Q7

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	M13-TZ2-P1-3(HL)
Question Text	Expand $(3 - 4x)⁵$ in ascending powers of $x$ , simplifying coefficients. Write the coefficient of $x³$ .
Total Mark	4
Correct Answer	-5760
Explanation	n/a
Mark Scheme	Attempt binomial expansion. $3^{5}+5\times 3^{4}\times (-4x)+\frac{5\times4}{2!}\times3^{3}\times(-4x)^{2}+\frac{5\times4\times3}{3!}\times3^{2}\times(-4x)^{3}+\frac{5\times4\times3\times2}{4!}\times3\times(-4x)^{4}+(-4x)^{5}$ $= 242-1620x+4320x^2-5760x^3+3840x^4-1024x^5$ Hence, the coefficient of $x^3$ is equal to $-5760$

Q8

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	M19/5/MATHL/HP1/ENG/TZ1/XX/2
Question Text	Determine sum of first three terms’ coefficients of $(1 - 2x)¹¹$ in ascending powers of $x$ , giving each term in its simplest form.
Total Mark	4
Correct Answer	-241
Explanation	n/a
Mark Scheme	Attempt binomial expansion. $\left ( 1-2x \right )^{11}$ $=1^{11}+{}^{11}C_1\cdot1^{10}\cdot(-2x)^{1}+{}^{11}C_2\cdot1^{9}\cdot(-2x)^{2}+{}^{11}C_3\cdot 1^{8}\cdot (-2x)^{3}+…$ $= 1-22x-220x^2-1320x^3+…$ Hence, the sum of the first three terms’ coefficients can be calculated as below. $1-22-220=-241$

Q9

Topic	1.4 Binomial Theorems and Combinatorics
Tag	Binomial theorem; Combination; Permutation; Pascal’s Triangle; Counting principle; Expansion; Factorial
Source	N18/5/MATHL/HP1/ENG/TZ0/XX/2
Question Text	A team of four is to be chosen from a group of five boys and five girls. (a) Find the number of different possible teams that could be chosen.
Total Mark	2
Correct Answer	210
Explanation	n/a
Mark Scheme	Use combinations. ${}^{10}C_4=\frac{10!}{4!6!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$
Question Text	(b) Find the number of different possible teams that could be chosen, given that the team must include at least one girl and at least one boy.
Total Mark	1
Correct Answer	200
Explanation	n/a
Mark Scheme	METHOD 1 The answer is the total number of teams minus the number of teams with all girls or all boys. Total number of teams: $210$ Number of teams with all boys: ${}^{5}C_4=5$ Number of teams with all girls: ${}^{5}C_4=5$ Hence, the number of different possible teams are $210-5-5=200$ METHOD 2 The answer is the total number of teams with 1 boy, 2 boys, and 3 boys. Number of teams with 1 boy: ${}^{5}C_1 \times {}^{5}C_3=50$ Number of teams with 2 boys: ${}^{5}C_2 \times {}^{5}C_2=100$ Number of teams with 3 boys: ${}^{5}C_3 \times {}^{5}C_1 = 50$ Hence, the number of different possible teams are $50+100+50=200$