Edexcel International A Level Maths · P2 · Pure Mathematics 2
The binomial expansion
28 free, high-quality past-paper-style questions on The binomial expansion for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 224 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-4-B-01P2 · ASEasy6 marks
Throughout this question n is a positive integer, and nCr = n!/(r!(n − r)!).
(a)
Use this definition to evaluate 10C3, showing your working.
(2)
(b)
Show that (n + 1)!/(n − 1)! = n(n + 1).
(2)
(c)
Given that nC2 = 28, find the value of n.
(2)
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Answer
Mark
Notes
Part (a)
10!/(3! 7!) = (10 × 9 × 8)/(3 × 2 × 1)
M1
120
A1
Part (b)
writes (n + 1)! = (n + 1)n(n − 1)!
M1
cancels (n − 1)! to leave n(n + 1), as required
A1
Part (c)
n(n − 1)/2 = 28, so n2 − n − 56 = 0
M1
(n − 8)(n + 7) = 0, so n = 8 (rejecting −7)
A1
Exam tip: Cancel before you multiply: 10!/7! is just 10 × 9 × 8. Writing the factorials out in full wastes time and invites slips.
Single terms of the expansion of (4 − x/2)7 can be found without writing out the whole expansion.
(a)
Find the coefficient of x3.
(3)
(b)
Find the term in x6, giving its coefficient as an exact fraction.
(2)
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Answer
Mark
Notes
Part (a)
uses 7C3(44)(−x/2)3
M1
35 × 256 × (−1/8)
A1
−1120
A1
Part (b)
7C6(4)(−x/2)6 = 7 × 4 × x6/64
M1
(7/16)x6
A1
Exam tip: A term asks for the power of x as well as the number; a coefficient asks for the number only. An even power of a negative quantity is positive.
Figure 1 shows rows 0 to 7 of Pascal's triangle. Three of the entries have been replaced by the letters p, q and r.
(a)
Find the values of p, q and r.
(3)
(b)
Use row 7 of Figure 1 to find the term in x2 in the expansion of (3 + x)7.
(2)
(c)
Find the sum of the entries in row 7, and explain, using the expansion of (1 + x)7, why this sum is a power of 2.
(1)
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Answer
Mark
Notes
Part (a)
p = 4 + 6 = 10
B1
q = 10 + 10 = 20
B1
r = 6 + 15 = 21
B1
Part (b)
third entry of row 7 with 35: 21 × 35x2
M1
5103x2 ft their r
A1
Part (c)
128: putting x = 1 in (1 + x)7 adds every coefficient, so the sum is 27 = 128
B1
Exam tip: Row n of Pascal's triangle starts with row 0 = 1, so the row for a seventh power is the one that starts 1, 7, … The entry with x2 is the third one along, not the second.
Figure 1 shows a cube of edge (5 + h) cm, where h > 0. Three cuts, one parallel to each pair of faces, divide every edge into a 5 cm part and an h cm part, so the cube is split into eight cuboids.
(a)
By finding the volume of each type of cuboid in Figure 1 and the number of cuboids of each type, show that the volume of the cube is (125 + 75h + 15h2 + h3) cm3.
(3)
(b)
State how the numbers of cuboids of each type are connected to the binomial expansion of (5 + h)3.
(1)
(c)
The edge of the cube is increased from 5 cm to 5.2 cm. Use part (a) to find the exact increase in volume.
(2)
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Answer
Mark
Notes
Part (a)
types: one 5 × 5 × 5, three 5 × 5 × h, three 5 × h × h, one h × h × h
1, 3, 3, 1 are the binomial coefficients 3C0, 3C1, 3C2, 3C3 (row 3 of Pascal's triangle)
B1
Part (c)
h = 0.2: 75(0.2) + 15(0.2)2 + 0.23
M1
= 15 + 0.6 + 0.008 = 15.608 cm3
A1
Exam tip: The increase in volume is everything except the 125 cm3 block. For a small h the 75h term does almost all the work, which is why the first terms of an expansion give good estimates.
Figure 1 shows a card on which the first terms of the binomial expansion of (2 + kx)n, in ascending powers of x, were written. Two of the coefficients are hidden by ink blots, but the signs shown on the card are correct.
(a)
Find the value of n.
(1)
(b)
Find the value of k, explaining how Figure 1 lets you choose its sign.
(2)
(c)
Find the two coefficients hidden by the ink blots.
(2)
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Answer
Mark
Notes
Part (a)
2n = 64, so n = 6
B1
Part (b)
6C2 × 24 × k2 = 240k2 = 60, so k2 = 1/4
M1
the x term is negative and 6 × 25k has the sign of k, so k = −1/2
A1
Part (c)
coefficient of x: 6 × 32 × (−1/2) = −96, so the blot hides 96 (allow 96 or −96) ft their k
B1
coefficient of x3: 20 × 8 × (−1/8) = −20, so the blot hides 20 (allow 20 or −20) ft their k
B1
Common error: Taking the square root of k2 = 1/4 and keeping only the positive value. Squared terms hide the sign of k; you need an odd power, here the x term, to decide it.
A machine is bought for £8000. Its value falls by the same percentage at the end of every year. Figure 1 shows its value at the start of each year, t = 0 to 6; three of the bars are labelled with their values, to the nearest penny.
(a)
Use Figure 1 to show that the value falls by 5% each year.
(1)
(b)
Find the first three terms, in ascending powers of x, of the binomial expansion of (1 − x)6.
(2)
(c)
Use your expansion with a suitable value of x to estimate the value of the machine when t = 6.
(2)
(d)
Explain why your estimate is larger than the value labelled on Figure 1.
(1)
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Answer
Mark
Notes
Part (a)
7600/8000 = 0.95, a fall of 5%
B1
Part (b)
6C2(−x)2 or equivalent structure
M1
1 − 6x + 15x2
A1
Part (c)
x = 0.05: 8000(1 − 0.3 + 0.0375)
M1
= £5900
A1
Part (d)
the next term, −20x3, is negative (worth −£20), so leaving it out makes the estimate too high
B1
Exam tip: A fall of 5% means multiplying by 0.95 = 1 − 0.05, so the bracket is (1 − x) with x = 0.05. Using x = 0.95 in the expansion gives nonsense.
In the binomial expansion of (1 + kx)n, where n is a positive integer and k is a constant, the coefficient of x is −15 and the coefficient of x2 is 90.
(a)
Write down two equations in n and k.
(2)
(b)
Find the value of n and the value of k.
(4)
(c)
Find the coefficient of x3.
(2)
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Answer
Mark
Notes
Part (a)
nk = −15
B1
n(n − 1)k2/2 = 90
B1
Part (b)
divides the second equation by the first, or substitutes k = −15/n into the second
M1
(n − 1)k/2 = −6, so (n − 1)k = −12; or 225(n − 1)/(2n) = 90
A1
solves the resulting linear equation for one unknown, e.g. subtracts from nk = −15 to get k
M1
k = −3, n = 5
A1
Part (c)
5C3(−3)3 = 10 × (−27)
M1
−270 ft from their n and k
A1
Exam tip: Dividing one coefficient equation by the other removes a power of the constant and part of the binomial coefficient at the same time, leaving a linear equation.
In this question a is a positive constant. Consider the binomial expansion of (2 + ax)7.
(a)
Find, in terms of a, the coefficient of x2 and the coefficient of x4.
(3)
(b)
Given that these two coefficients are equal, find the exact value of a, giving your answer in the form p√q, where p is a rational number and q is an integer.
(3)
(c)
Find the value of these equal coefficients, as an exact fraction.
(1)
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Answer
Mark
Notes
Part (a)
uses 7C2(25)a2 or 7C4(23)a4
M1
672a2
A1
280a4
A1
Part (b)
280a4 = 672a2 and divides by a2 ≠ 0
M1
a2 = 672/280 = 12/5
A1
a = (2/5)√15
A1
Part (c)
672 × 12/5 = 8064/5
B1
Common error: Cancelling a2 without comment. You may divide by it only because a ≠ 0; a = 0 would also make the coefficients equal, but it is excluded here.
The expansion of (1 − x/5)8 is used to approximate a power of a number slightly less than 1.
(a)
Find the first four terms, in ascending powers of x, giving each coefficient as an exact fraction.
(4)
(b)
State the value of x that you would use to estimate 0.9988 from this expansion.
(1)
(c)
Use your expansion to estimate 0.9988, giving your answer to 6 decimal places.
(2)
(d)
Explain why the same four terms would not give a useful estimate of 0.58.
(1)
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Answer
Mark
Notes
Part (a)
correct structure, e.g. 8C2(−x/5)2
M1
1 − (8/5)x
A1
+ (28/25)x2
A1
− (56/125)x3
A1
Part (b)
1 − x/5 = 0.998, so x = 0.01
B1
Part (c)
substitutes their x into their four terms: 1 − 0.016 + 0.000112 − 0.000000448
M1
0.984112
A1
Part (d)
it needs x = 2.5, which is not small, so the terms left out are not negligible (the four terms give −3, nowhere near 0.0039)
B1
Exam tip: Show the equation that fixes x, such as 1 − x/5 = 0.998. Substituting the number you want to raise to the power (0.998) for x is a common slip.
Find the first four terms, in ascending powers of x, of the binomial expansion of (1 − 3x)6.
(3)
(b)
Find the coefficient of x in the expansion of f(x).
(1)
(c)
Find the coefficient of x2 and the coefficient of x3 in the expansion of f(x).
(3)
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Answer
Mark
Notes
Part (a)
correct structure, e.g. 6C3(−3x)3
M1
two of −18x, 135x2, −540x3 correct
A1
1 − 18x + 135x2 − 540x3
A1
Part (b)
2 × (−18) + 1 × 1 = −35
B1
Part (c)
identifies both pairs for one power, e.g. 2 × 135 and 1 × (−18) for x2
M1
x2: 252
A1
x3: 2 × (−540) + 135 = −945
A1
Exam tip: Line the two brackets up and ask, for each power you need, which pairs of terms multiply to give it. Each power of x here comes from exactly two pairs.
In the binomial expansion of (a + x)n, where a and n are positive integers, the first three coefficients are shown in the bar chart in Figure 1. The bar for the coefficient of x3 has been left out.
(a)
Use the first two bars to show that n = 3a.
(2)
(b)
Use the third bar to find the value of a and the value of n.
(3)
(c)
Find the coefficient of x3, which belongs to the missing bar.
(2)
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Answer
Mark
Notes
Part (a)
an = 64 and nan − 1 = 192
M1
dividing: n/a = 3, so n = 3a
A1
Part (b)
½n(n − 1)an − 2 = 240
M1
divides by the x coefficient: (n − 1)/(2a) = 240/192 = 5/4
M1
4(3a − 1) = 10a, so a = 2, n = 6 (and 26 = 64 checks)
A1
Part (c)
6C3 × 23
M1
160 ft their a, n
A1
Exam tip: Dividing one coefficient by the one before it cancels most of the powers of a and most of the binomial coefficient, leaving a simple ratio such as (n − 1)/(2a).
Figure 1 shows part of the curve with equation y = (a + bx)4, where a and b are positive constants. The curve passes through the points P(0, 81) and Q(−3, 81) and touches the x-axis at M(−1.5, 0).
(a)
Use the points P and M to find the value of a and the value of b.
(3)
(b)
Expand (a + bx)4 fully, simplifying each term.
(3)
(c)
Use your expansion, with a suitable value of x, to find the exact value of 3.024.
Exam tip: A curve that touches the x-axis without crossing it has a repeated root. Here the whole bracket is zero there, so a + bx = 0 at M gives b straight away. Q is a free check: (3 − 6)4 = 81.
Figure 1 shows a grid being used to find the first terms of the expansion of (3 − 2x)(1 + x/2)6 in ascending powers of x. The top row holds the first four terms of the expansion of (1 + x/2)6, two of which are written as P and Q. Each cell holds the product of its row heading and its column heading.
(a)
Find P and Q, giving each in its simplest form.
(3)
(b)
Write down the entries A, B and C.
(2)
(c)
Hence find the first three terms of the expansion of (3 − 2x)(1 + x/2)6, and show that it has no term in x3.
(3)
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Answer
Mark
Notes
Part (a)
P = 6C2(x/2)2 or Q = 6C3(x/2)3
M1
P = (15/4)x2
A1
Q = (5/2)x3
A1
Part (b)
A = (45/4)x2 ft
B1
B = −(15/2)x3 and C = (15/2)x3 ft
B1
Part (c)
adds the cells with equal powers of x, e.g. 9x − 2x
M1
3 + 7x + (21/4)x2
A1
x3: C + B = (15/2)x3 − (15/2)x3 = 0, so there is no x3 term
A1
Exam tip: In a multiplication grid the cells with the same power of x lie on a diagonal running from top right to bottom left. Collect each diagonal; a cell whose power is too high can be left empty.
The first three terms of the binomial expansion of (1 + x)6 are used as an approximation, so that (1 + x)6 ≈ 1 + 6x + 15x2. The error in this approximation is e(x) = (1 + x)6 − (1 + 6x + 15x2). Figure 1 shows the graph of e(x) for 0 ≤ x ≤ 0.24.
(a)
Find the term in x3 of the expansion of (1 + x)6, and hence write down an approximate expression for e(x) when x is small.
(2)
(b)
Use your answer to part (a) to estimate e(0.1), and check your estimate against Figure 1.
(2)
(c)
Use Figure 1 to find the largest value of x for which the error in the approximation is less than 0.05.
(2)
(d)
Without drawing on Figure 1 further, decide whether 1 + 6(0.15) + 15(0.15)2 is within 0.05 of 1.156. Give a reason.
(2)
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Answer
Mark
Notes
Part (a)
6C3x3 = 20x3
B1
e(x) ≈ 20x3, since the remaining terms have higher powers of x
B1
Part (b)
20 × 0.13 = 0.02
B1
the graph at x = 0.1 reads about 0.02 (allow 0.015 to 0.025), which agrees
B1
Part (c)
reads where the curve crosses the line e = 0.05
M1
x ≈ 0.13 (allow 0.125 to 0.137)
A1
Part (d)
0.15 is larger than their answer to (c), and e(x) increases with x
M1
so the error is more than 0.05: not within 0.05 (the actual error is about 0.076)
A1
Exam tip: The first term left out is a good guide to the error only while x is small. At x = 0.2 the guide 20x3 gives 0.16, but the graph shows nearly 0.19, because the x4 and higher terms are no longer negligible.
Figure 1 shows, for 0 ≤ x ≤ 1, the curve C with equation y = (1 + kx)5, where k is a positive constant, together with the line L, whose equation is formed from the first two terms of the binomial expansion of (1 + kx)5, and the dashed curve Q, whose equation is formed from the first three terms. The line L passes through the labelled point (0.5, 2).
(a)
Write down the equation of L in terms of k, and use Figure 1 to show that k = 0.4.
(3)
(b)
Find the equation of Q.
(2)
(c)
Use the equation of Q with a suitable value of x to estimate 1.045.
(2)
(d)
Using Figure 1, explain why Q would not give a good estimate of 1.45.
(1)
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Answer
Mark
Notes
Part (a)
L: y = 1 + 5kx
B1
gradient of L from (0, 1) and (0.5, 2) is 2
M1
5k = 2, so k = 0.4
A1
Part (b)
third term 5C2(0.4x)2
M1
y = 1 + 2x + 1.6x2
A1
Part (c)
1 + 0.4x = 1.04, so x = 0.1
M1
1 + 0.2 + 0.016 = 1.216
A1
Part (d)
1.45 needs x = 1, where Q (4.6) is well below C (about 5.4): kx = 0.4 is not small, so the left-out terms matter
B1
Exam tip: The first two terms of (1 + kx)n give a straight line through (0, 1) with gradient nk. Each extra term bends the approximation closer to the curve, but only near x = 0.
Figure 1 shows the curve with equation y = (3 + x)4 + (3 − x)4 and the line y = 272, which meet at the points A and B. The lowest point of the curve is (0, 162).
(a)
Expand (3 + x)4 fully, simplifying each term.
(3)
(b)
Hence show that y = 162 + 108x2 + 2x4, and use this form to explain why the curve is symmetrical about the y-axis and why its lowest point is (0, 162).
(3)
(c)
Find the x-coordinates of A and B.
(3)
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Answer
Mark
Notes
Part (a)
correct structure, e.g. 4C2(32)x2
M1
at least three terms correct
A1
81 + 108x + 54x2 + 12x3 + x4
A1
Part (b)
(3 − x)4 = 81 − 108x + 54x2 − 12x3 + x4, by replacing x with −x
M1
adding, the odd powers cancel: y = 162 + 108x2 + 2x4
A1
only even powers, so x and −x give the same y; and 108x2 + 2x4 ≥ 0 with equality only at x = 0
B1
Part (c)
162 + 108x2 + 2x4 = 272, so x4 + 54x2 − 55 = 0
M1
(x2 + 55)(x2 − 1) = 0, a quadratic in x2
M1
x2 = −55 has no real roots, so x = −1 at A and x = 1 at B
A1
Common error: Writing (3 − x)4 with every term negative. Only the odd powers of −x are negative: (−x)2 = x2 and (−x)4 = x4.
so the sum is (n + 1)!/((r + 1)!(n − r)!) = n + 1Cr + 1, with a conclusion
A1
Part (b)
two neighbouring entries in row n add to give the entry between them in row n + 1
B1
Part (c)
nC3/nC2 = (n − 2)/3, or writes both as products in n
M1
(n − 2)/3 = 2
M1
n = 8
A1
Part (d)
28k2 = 56k3
M1
k = ½
A1
Exam tip: In a factorial proof, aim for the common denominator (r + 1)!(n − r)!. Multiplying top and bottom by the missing factor is the only real step; everything else is tidying.
An expansion of (2 − x/4)10 is used to estimate a power of 1.99, and the size of the first neglected term is used to judge how accurate that estimate is.
(a)
Find the first four terms, in ascending powers of x, giving each term in its simplest form.
(4)
(b)
Use your expansion, with a suitable value of x, to find an estimate of 1.9910, giving your answer to 2 decimal places.
(3)
(c)
Find the term in x4 of the expansion, and show that at your value of x it is less than 0.001. Explain what this tells you about your answer to part (b).
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
correct structure, e.g. 10C2(28)(−x/4)2
M1
1024 − 1280x
A1
+ 720x2
A1
− 240x3
A1
Part (b)
2 − x/4 = 1.99, so x = 0.04
B1
1024 − 51.2 + 1.152 − 0.01536
M1
= 973.93664 = 973.94 (2 d.p.)
A1
Part (c)
10C4(26)(−x/4)4 = 52.5x4
M1
52.5 × 0.044 = 0.000134… < 0.001
A1
the first neglected term (and the even smaller ones after it) cannot change the second decimal place, so 973.94 is reliable to 2 d.p.
B1
Exam tip: The size of the first term you left out is a quick guide to the accuracy of a binomial estimate, as long as the terms are shrinking fast.
In this question k is a constant. Let C denote the coefficient of x2 in the expansion of (2 − 5x)(1 + kx)6. Figure 1 shows the graph of C against k. The line C = 60 meets the graph at Q and P, and M is the lowest point of the graph.
(a)
Show that C = 30k2 − 30k.
(3)
(b)
Find the coordinates of M, and so state the least possible value of the coefficient of x2.
(2)
(c)
Find the value of k at P and the value of k at Q.
(3)
(d)
For the value of k at P, find the coefficient of x3 in the expansion of (2 − 5x)(1 + kx)6.
(3)
(e)
Using Figure 1, or otherwise, state the set of values of k for which the coefficient of x2 is negative.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
(1 + kx)6 = 1 + 6kx + 15k2x2 + …
B1
x2 terms: 2 × 15k2x2 and (−5x) × 6kx
M1
C = 30k2 − 30k, as required
A1
Part (b)
30(k − ½)2 − 7.5, or differentiates and solves 60k − 30 = 0
M1
M(0.5, −7.5); least value −7.5
A1
Part (c)
30k2 − 30k = 60, so k2 − k − 2 = 0
M1
(k − 2)(k + 1) = 0
M1
P: k = 2; Q: k = −1
A1
Part (d)
x3 terms: 2 × 6C3k3 and (−5) × 6C2k2
M1
40k3 − 75k2
A1
320 − 300 = 20 ft their k
A1
Part (e)
0 < k < 1
B1
Exam tip: Two pairs of terms give x2: the constant 2 with the x2 term, and −5x with the x term. Write both pairs down before simplifying, and keep the minus sign with the 5x.
Figure 1 shows part of the curve with equation y = (p + qx)5, where p and q are constants. The curve meets the y-axis at A(0, 243) and the x-axis at B(3, 0), and passes through C(2, 1).
(a)
Use the points A and B to find the value of p and the value of q, and verify that C lies on the curve.
(3)
(b)
Find the first four terms, in ascending powers of x, of the binomial expansion of (p + qx)5.
(4)
(c)
Hence find the first three terms, in ascending powers of x, of the expansion of (2 + x)(p + qx)5.
(3)
(d)
A student substitutes x = 2 into the four terms from part (b) to estimate the y-coordinate of C. Find the value obtained and explain, with reference to Figure 1, why it is so far from the true value.
(2)
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Answer
Mark
Notes
Part (a)
p5 = 243, so p = 3
B1
3 + 3q = 0
M1
q = −1, and (3 − 2)5 = 1 checks C
A1
Part (b)
correct structure, e.g. 5C2(33)(−x)2
M1
243 − 405x
A1
+ 270x2
A1
− 90x3, giving 243 − 405x + 270x2 − 90x3
A1
Part (c)
multiplies out, e.g. x2: 2 × 270 and 1 × (−405)
M1
486 − 567x
A1
+ 135x2
A1
Part (d)
243 − 810 + 1080 − 720 = −207
B1
the true value at C is 1; x = 2 is not small (the terms grow instead of shrinking), so the terms left out are not negligible
B1
Common error: Losing the alternating signs. With q = −1 every odd power of −x is negative, so the signs go +, −, +, −. A sign slip in the x term spoils every later product in part (c).
Two savings accounts, A and B, each pay compound interest at a fixed rate once a year. No money is added or taken out. Figure 1 shows the balance of each account at the end of each year t = 0 to 10; the balances for t = 0 and t = 1 are labelled. Do not use the power key of a calculator in this question.
(a)
Use Figure 1 to find the annual interest rate for each account.
(2)
(b)
Use the first three terms of the binomial expansion of (1 + x)6 to estimate the balance of each account when t = 6, and state which account these estimates suggest is larger.
(4)
(c)
Find, for each account, the size of the next term of the expansion, and explain why your conclusion in part (b) can be trusted.
(3)
(d)
Figure 1 suggests that the order of the two balances changes between t = 6 and t = 7. Use the first three terms of the expansion of (1 + x)7 to support this.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
A: 2100/2000 = 1.05, so 5%
B1
B: 2448/2400 = 1.02, so 2%
B1
Part (b)
(1 + x)6 ≈ 1 + 6x + 15x2
M1
A: 2000(1 + 0.3 + 0.0375) = £2675
A1
B: 2400(1 + 0.12 + 0.006) = £2702.40
A1
B is larger
A1
Part (c)
next term 20x3: A: 2000 × 20 × 0.053 = £5
M1
B: 2400 × 20 × 0.023 = £0.384
A1
later terms are smaller still, so the true value of A is only a few pounds above £2675, while the gap is about £27: B is larger
A is now larger by about £49, far more than the size of the terms left out, so the order changes
A1
Exam tip: Every term of (1 + x)n is positive when x > 0, so a truncated expansion always underestimates a growing balance. Compare the gap between the estimates with the size of the next term before drawing a conclusion.
Figure 1 shows three consecutive entries, 120, 210 and 252, from row n of Pascal's triangle. They are nCr, nCr + 1 and nCr + 2. The entries X and Y in row n + 1 sit below them as shown.
(a)
Show that nCr + 1 ÷ nCr = (n − r)/(r + 1).
(2)
(b)
Hence find the value of n and the value of r.
(4)
(c)
Find X and Y.
(2)
(d)
In the binomial expansion of (2 + x)n, using your value of n, find the ratio of the coefficient of x5 to the coefficient of x4, in its simplest form.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
writes the quotient with factorials: n!/((r + 1)!(n − r − 1)!) × r!(n − r)!/n!
M1
cancels using (r + 1)! = (r + 1)r! and (n − r)! = (n − r)(n − r − 1)!, giving (n − r)/(r + 1)
A1
Part (b)
(n − r)/(r + 1) = 210/120 = 7/4
M1
(n − r − 1)/(r + 2) = 252/210 = 6/5
M1
4n = 11r + 7 and 5n = 11r + 17
A1
n = 10, r = 3
A1
Part (c)
X = 120 + 210 = 330
B1
Y = 210 + 252 = 462
B1
Part (d)
coefficient of x4: 210 × 26, or of x5: 252 × 25
M1
13440 and 8064
A1
3 : 5
A1
Exam tip: Ratios of neighbouring binomial coefficients are much easier to handle than the coefficients themselves: the factorials cancel to a single fraction, and two ratios give two linear equations.
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