Binomial theorem and counting principles – IB Maths exam questions with mark scheme

1CoreNon-calculator3 marks

Expand $(x+2)^4$, giving your answer in descending powers of $x$.

Uses binomial coefficients $1,4,6,4,1$ with powers of 2M1
At least three correct termsA1
$x^4+8x^3+24x^2+32x+16$A1
Answer: $x^4+8x^3+24x^2+32x+16$
Examiner tip: Remember $2^2=4$ and $2^3=8$ are multiplied by the coefficient, not replaced by it.

2StandardNon-calculator4 marks

Find the constant term in the expansion of $\left(x^2+\dfrac{2}{x}\right)^9$.

General term $\binom9r(x^2)^{9-r}\left(\dfrac2x\right)^r=\binom9r2^rx^{18-3r}$M1
Sets $18-3r=0$M1
$r=6$A1
$\binom96\times2^6=84\times64=5376$A1
Answer: $5376$
Examiner tip: Combine the powers of $x$ first, then set the exponent to zero.

3CoreNon-calculatorHL only4 marks

A club has 8 members.

(a) A committee of 3 members is chosen. Find the number of possible committees. [2]

(b) Instead, a president, a vice-president and a secretary are chosen from the 8 members. Find the number of ways this can be done. [2]

$\binom83$ (or $\dfrac{8\times7\times6}{3!}$)M1
$56$A1
$8\times7\times6$ or $^8P_3$M1
$336$A1
Answer: (a) $56$ (b) $336$
Examiner tip: Order matters in (b) but not in (a): that is why (b) is $3!$ times larger.

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4StandardCalculatorHL only6 marks

Consider the 10 letters of the word STATISTICS.

(a) Find the number of different arrangements of all 10 letters. [3]

(b) Find the number of these arrangements in which the three letters S are next to each other. [3]

$\dfrac{10!}{\cdots}$ (correct numerator)M1
Divides by $3!\,3!\,2!$ for repeated S, T, IM1
$50\,400$A1
Treats SSS as one object, giving 8 objectsM1
$\dfrac{8!}{3!\,2!}$ (repeats of T and I)M1
$3360$A1
Answer: (a) $50\,400$ (b) $3360$
Examiner tip: Divide by the factorial of every repeated letter, including those in the block method.

5StandardNon-calculatorHL only4 marks

A team of 4 is chosen from 6 men and 5 women. The team must contain at least 2 women. Find the number of possible teams.

Valid method: cases (2, 3, 4 women) or total minus (0 or 1 woman)M1
Complement: $\binom{11}4=330$ and $\binom64=15$ (no women); or direct: $\binom52\binom62=150$ (2 women)A1
Complement: $\binom51\binom63=100$ (one woman); or direct: $\binom53\binom61=60$ and $\binom54=5$A1
$330-15-100=215$ or $150+60+5=215$A1
Answer: $215$
Examiner tip: 'At least' questions are often quicker by complement; check by the direct method if time allows.

6HardNon-calculatorHL only4 marks

How many four-digit numbers greater than 4000 can be formed from the digits $1,2,3,4,5,6,7$ with no digit used more than once, if the number must be even?

Splits into cases by parity of the first digitM1
First digit even (4 or 6): $2\times2\times5\times4=80$A1
First digit odd (5 or 7): $2\times3\times5\times4=120$A1
$200$A1
Answer: $200$
Examiner tip: Fill the most restricted positions (first and last digits) first, and watch the overlap when both must be even.

7CoreNon-calculator5 marks

The diagram shows the first seven rows of Pascal's triangle, from $n=0$ to $n=6$. Two entries in the row $n=6$ have been replaced by $p$ and $q$.

(a) Write down the value of $p$ and the value of $q$. [2]

(b) Hence, or otherwise, find the term in $x^2$ in the expansion of $(2-x)^6$. [3]

(a) $p=15$A1
(a) $q=20$A1
(b) Recognises the correct term: $\binom{6}{2}$ (or $p$) with $2^4$ and $(-x)^2$, e.g. $15\times2^4\times(-x)^2$. Accept $\binom{6}{4}$; allow a sign error in $(-x)$ for this mark onlyM1
(b) Correct unsimplified term, e.g. $15(16)(x^2)$ or $15\times16x^2$A1
(b) $240x^2$ (the question asks for the term, so $240$ alone scores A0 for this mark; accept $240x^2$ picked out from a full correct expansion)A1
Answer: (a) $p=15,\ q=20$ (b) $240x^2$
Examiner tip: Students often use the coefficient for the wrong power of 2: the $x^2$ term pairs $\binom{6}{2}$ with $2^{4}$, not $2^{2}$.

8StandardNon-calculator7 marks

In the expansion of $(1+kx)^n$, where $n\in\mathbb{Z}^+$ and $k\in\mathbb{R}$, the coefficient of $x$ is $20$ and the coefficient of $x^2$ is $150$.

(a) Find the value of $n$ and the value of $k$. [5]

(b) Find the coefficient of $x^3$. [2]

(a) $nk=20$A1
(a) $\binom{n}{2}k^2=150$, i.e. $\dfrac{n(n-1)}{2}k^2=150$A1
(a) Valid attempt to eliminate one variable, e.g. substitutes $k=\dfrac{20}{n}$, or divides to obtain $\dfrac{n(n-1)k^2}{n^2k^2}=\dfrac{300}{400}$M1
(a) $n=4$ (from $\dfrac{n-1}{n}=\dfrac34$ or $400(n-1)=300n$)A1
(a) $k=5$A1
(b) $\binom{4}{3}\times5^3$ (or $4\times125$); follow through their $n$ and $k$M1
(b) $500$. Accept $500x^3$A1
Answer: (a) $n=4,\ k=5$ (b) $500$
Examiner tip: Many students forget to square $k$ in the $x^2$ coefficient: it is $\binom{n}{2}k^2$, not $\binom{n}{2}k$.

9HardCalculatorHL only7 marks

Let $f(x)=\sqrt{4+x}$.

(a) Find the first three terms in the binomial expansion of $f(x)$ in ascending powers of $x$. [4]

(b) State the set of values of $x$ for which this expansion is valid. [1]

(c) Use your answer to part (a) with a suitable value of $x$ to find an approximation for $\sqrt{4.1}$. Find the percentage error in this approximation. [2]

(a) Takes out a factor of 4: $\sqrt{4+x}=2\left(1+\dfrac{x}{4}\right)^{\frac12}$ (or $4^{\frac12}(\ldots)$)M1
(a) Attempt at extended binomial expansion with $n=\tfrac12$: $1+\tfrac12\left(\tfrac{x}{4}\right)+\dfrac{\tfrac12\left(-\tfrac12\right)}{2!}\left(\tfrac{x}{4}\right)^2$M1
(a) Two of the three terms $2,\ \tfrac{x}{4},\ -\tfrac{x^2}{64}$ correctA1
(a) $2+\dfrac{x}{4}-\dfrac{x^2}{64}$A1
(b) $|x|<4$ (accept $-4A1
(c) Substitutes $x=0.1$ into their expansion to obtain $2.02484375$ (accept $2.02484$ or 3 s.f. $2.02$ only if more figures seen)A1
(c) Percentage error $=\left|\dfrac{2.02484375-\sqrt{4.1}}{\sqrt{4.1}}\right|\times100=9.50\times10^{-5}\ \%$ (accept $0.0000950\%$; accept $9.5\times10^{-5}\%$)A1
Answer: (a) $2+\dfrac{x}{4}-\dfrac{x^2}{64}$ (b) $|x|<4$ (c) $2.02484375$; percentage error $9.50\times10^{-5}\%$
Examiner tip: The expansion of $(1+y)^n$ needs a leading 1: factor out $4^{1/2}=2$ first, and the validity condition then becomes $|x/4|<1$, i.e. $|x|<4$.

10StretchNon-calculatorHL only8 marks

Four adults and three children sit in a row of seven seats. All seven people are different.

(a) Find the number of possible arrangements in which no two children sit next to each other. [3]

(b) Find the number of possible arrangements in which a child sits in each of the two end seats. [2]

A different group contains $n$ people, where $n\geq3$. The number of ways of choosing 3 of these people is four times the number of ways of choosing 2 of them.

(a) Arranges the adults first, $4!=24$, and identifies 5 gaps (including the two ends) for the childrenM1
(a) Children placed in gaps in $5\times4\times3=60$ ways (accept $^5P_3$, or $\binom{5}{3}\times3!$)M1
(a) $1440$A1
(b) Children at the ends in $3\times2=6$ ways, multiplied by $5!$ for the remaining five peopleM1
(b) $720$A1
(c) Forms equation $\binom{n}{3}=4\binom{n}{2}$, i.e. $\dfrac{n(n-1)(n-2)}{6}=4\times\dfrac{n(n-1)}{2}$M1
(c) Cancels $n(n-1)$ (valid as $n\geq3$) to obtain $\dfrac{n-2}{3}=4$ or $n-2=12$A1
(c) $n=14$. Award A0 if $n=0$ or $n=1$ are also given as final answersA1
Answer: (a) $1440$ (b) $720$ (c) $n=14$
Examiner tip: For 'no two together', place the unrestricted people first and slot the others into the gaps; subtracting only the 'all three together' case from $7!$ gives the wrong answer.

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