Mathematics 9709/53 — October/November 2024

Walkthrough

We are told that 30% of residents own an electric car, so for any single resident the probability of owning one is $p = 0.3$ and the probability of not owning one is $q = 1 - 0.3 = 0.7$.

For three independently chosen residents, the probability that all three own is $p^3 = 0.3^3 = 0.027$ because the choices are independent and we multiply the individual probabilities. Similarly, the probability that none owns one is $q^3 = 0.7^3 = 0.343$.

The two events "all three own" and "none owns" cannot both happen at the same time, so they are mutually exclusive. Therefore the probability of either happening is their sum: $0.027 + 0.343 = 0.37$.

Key Takeaways

This question tests two core laws of probability: the multiplication law for independent events (used to find the probability that several independent outcomes all occur) and the addition law for mutually exclusive events (used to combine mutually exclusive outcomes). An equally valid alternative approach is to subtract from 1 the probabilities of getting exactly one or exactly two owners.

Common Mistakes

A frequent error is to multiply the two probabilities, computing $0.3^3 \times 0.7^3$, because this treats the situation as if both conditions must hold simultaneously, which is not the case. Another common mistake is to write $0.3 + 0.7 = 1$ instead of $0.3^3 + 0.7^3$. Students also sometimes forget that the three selections are independent and fail to raise each probability to the third power.

Things to Be Careful About

The two events "all three own" and "none owns" are mutually exclusive, so the addition law applies directly. Be careful to cube each probability correctly: $0.3^3 = 0.027$ and $0.7^3 = 0.343$. The mark scheme also accepts the complement form

which is a useful check that the answer $0.37$ is correct.

Walkthrough

The variable $X$, the number of residents in a sample of 125 who own an electric car, follows a binomial distribution with $n = 125$ and $p = 0.3$.

To use the normal approximation, we first check the conditions: $np = 125 \times0.3 = 37.5 \geq 5$ and $n(1-p) = 125 \times0.7 = 87.5 \geq 5$, so the approximation is valid.

The mean and variance of the binomial distribution are:

so the standard deviation is $\sigma = \sqrt{26.25} \approx 5.123$.

Because the binomial is discrete, the event $X > 45$ must be adjusted by a continuity correction of $0.5$. In the continuous normal model, the boundary of the event becomes $45.5$, so we compute $P(X > 45.5)$.

We then standardise:

and since we want the upper tail probability,

Key Takeaway

This question tests the normal approximation to the binomial distribution, including checking the validity conditions ($np \geq5$ and $n(1-p) \geq5$), computing the binomial mean and variance, applying a continuity correction, and using the standard normal table. The key idea is that a discrete binomial can be modelled by a continuous normal distribution when $n$ is large, provided we account for the discreteness with a half-unit correction.

Common Mistakes

Forgetting the continuity correction is the most common error; some students use $45$ directly instead of $45.5$. Another frequent mistake is to divide by the variance $26.25$ instead of the standard deviation $\sqrt{26.25} \approx5.123$ when standardising. Some students also use the wrong tail, giving a probability greater than $0.5$; the correct answer must be below $0.5$ because $45$ is well above the mean of $37.5$.

Things to Be Careful About

Write the mean as $37.5$ and the variance as $26.25$ (or $26\frac{1}{4}$). The standard deviation is $\sqrt{26.25} \approx5.12$, sometimes written $\frac{\sqrt{105}}{2}$. Use the positive sign in the standardising formula because the boundary $45.5$ is above the mean. The final answer should satisfy $0 \leq p \leq1$ and, since the event is in the upper tail, $p < 0.5$; the mark scheme accepts $0.0592 \leq p \leq0.0593$.

Walkthrough

The red die has faces 1, 1, 1, 2, 2, 2, so it gives only two possible scores: 1 and 2, each with probability $\frac{1}{2}$. The blue die has faces 1, 1, 2, 2, 3, 3, so it gives scores 1, 2 and 3, each with probability $\frac{1}{3}$.

To find the possible values of $X$, multiply every red score by every blue score:

• $1 \times 1 = 1$
• $1 \times 2 = 2$ and $2 \times 1 = 2$
• $1 \times 3 = 3$
• $2 \times 2 = 4$
• $2 \times 3 = 6$

So the possible products are $1, 2, 3, 4, 6$. Products can occur in different ways; for example, $X=2$ can happen either as red 1 with blue 2, or red 2 with blue 1. Since the dice are independent and fair, multiply the individual score probabilities and add together any cases giving the same product.

The completed table lists each possible value of $X$ and its probability. The probabilities should sum to 1, which acts as a useful check.

Key Takeaways

A probability distribution table gives all possible values of a discrete random variable and their probabilities. Every possible value must be included, and the probabilities must sum to 1.

For independent events, probabilities are multiplied. When two different outcomes produce the same value of the random variable, their probabilities are added.

Common Mistakes

• Forgetting that $X = 2$ can occur in two different ways: red 1 with blue 2, or red 2 with blue 1.
• Omitting one of the possible values such as 4 or 6, or incorrectly including a value such as 5.
• Listing probabilities that do not sum to 1, which indicates a missing or incorrect case.
• Using the score on one die as the probability instead of counting the labelled faces.

Things to Be Careful About

• The possible values of $X$ are not consecutive; there is no product 5.
• The two dice are different, so the combinations red 1/blue 2 and red 2/blue 1 are distinct outcomes.
• The mark scheme allows the table values to be unsimplified, but the simplified fractions are clearer and should always sum to 1.
• If an extra $x$-value is included, its probability should be stated as 0.

Walkthrough

The expectation of a discrete random variable is its weighted average, where each possible value is multiplied by its probability. From part (a), the possible products are $1, 2, 3, 4, 6$ with probabilities $\frac{1}{6}, \frac{1}{3}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}$.

Write the expectation as

Substitute the probabilities, convert all terms to sixths, and add. The result is $\frac{18}{6} = 3$.

Key Takeaways

Expectation measures the long-run average value of a random variable. For a discrete random variable, it is found by summing value times probability over all possible values.

Using the probability distribution table makes the calculation straightforward: multiply each $x$ by its $P(X=x)$ and add the results.

Common Mistakes

• Using probabilities that do not sum to 1, which gives an invalid weighted average.
• Forgetting to multiply each $x$ by its probability, and instead just averaging the possible values.
• Arithmetic errors when working with different denominators.

Things to Be Careful About

• Use exact fractions rather than decimals to avoid rounding errors.
• The answer is exactly 3 because the probabilities are exact.
• If the table from part (a) were incorrect, the expectation could be follow-through marked, provided the probabilities used are valid and sum to 1.

Walkthrough

The random variable $X$ represents the height of an adult male and is modelled as $X \sim N(176, 4.8^2)$. To use the standard normal tables we must convert the problem into one about the standard normal variable $Z$ by subtracting the mean and dividing by the standard deviation:

Substituting $X = 170$ gives $z = \frac{170 - 176}{4.8} = -1.25$, so we need $P(Z > -1.25)$. The standard normal table typically gives $\Phi(z) = P(Z < z)$ for positive $z$ only, but by the symmetry of the bell curve, $P(Z > -1.25) = P(Z < 1.25) = \Phi(1.25)$. Reading from the table, $\Phi(1.25) = 0.8944$, so the probability is approximately $0.894$.

Key Takeaways

• The standardisation formula $Z = \frac{X - \mu}{\sigma}$ converts any normal variable to the standard normal.
• Because the normal distribution is symmetric, $P(Z > -a) = \Phi(a) = P(Z < a)$ for any $a > 0$.
• The Cambridge tables (or calculators) give the area to the left of $z$, so a "greater than" probability must be converted using symmetry or by taking $1 - \Phi(z)$.

Common Mistakes

• Forgetting to invert the answer when the question asks for $P(X > k)$ and reading $\Phi(-1.25)$ directly, which would give an answer less than $0.5$.
• Using $\sigma^2$ instead of $\sigma$ in the standardisation, e.g. dividing by $4.8^2 = 23.04$ instead of $4.8$.

Things to Be Careful About

• Make sure the standard deviation $4.8$ is used, not the variance.
• The probability of being greater than a value below the mean is greater than $0.5$; the answer should look "right".
• The mark scheme accepts any value in the range $0.89435 \leq p \leq 0.8944$.

Walkthrough

We already know from part (a) that $P(X > 170) = 0.8944$, so its complement $P(X < 170) = 1 - 0.8944 = 0.1056$. This is the area under the curve to the left of $170$.

The question states that $60\%$ of adult males have heights between $170$ and $k$ cm, which is a middle slice of the distribution. Adding this middle slice to the left-hand tail tells us the total area to the left of $k$:

Now we need the value of $k$ that produces exactly this cumulative probability. Standardising gives the equation

Using inverse normal tables, $\Phi^{-1}(0.7056) \approx 0.541$ (in particular, $\Phi(0.540) = 0.7054$ and $\Phi(0.541) = 0.7057$, so $0.541$ is the closest entry). Finally we undo the standardisation by multiplying by $4.8$ and adding $176$:

The answer makes intuitive sense: since the distribution is centred at $176$ and $60\%$ of the data lie between $170$ and $k$, the cut-off $k$ must be a few centimetres above the mean.

Key Takeaways

• An interval probability $P(a < X < b)$ can be written as $P(X < b) - P(X < a)$.
• Working backwards from a given probability to a value of $X$ requires the inverse normal function $\Phi^{-1}$.
• Always undo the standardisation by $X = \mu + z\sigma$ to recover the value in original units.

Common Mistakes

• Using $z = \Phi^{-1}(0.6)$ instead of $\Phi^{-1}(0.7056)$: this would be the cut-off if the question were "$60\%$ have height less than $k$", not "between $170$ and $k$".
• Forgetting to add the lower-tail area $0.1056$ to $0.6$ before taking the inverse.
• Using $z = 0.541$ from a calculator without showing the standardisation step that led to it.

Things to Be Careful About

• The mark scheme allows the $z$-value in the range $0.540 < z \leq 0.541$, so any answer in $k = 178.5$ to $k = 178.6$ is acceptable.
• The mark scheme specifically forbids substituting $4.8^2$ or $\sqrt{4.8}$ for the standard deviation, and no continuity correction is needed because heights are modelled as continuous.

Walkthrough

First, we identify the upper class boundaries for each height interval. Since the heights are measured to the nearest cm, a value in the class 10–19 actually represents heights from 9.5 up to 19.5. Thus, the upper boundaries are 19.5, 29.5, 39.5, 44.5, 49.5, 54.5, and 59.5. We also start at the lower boundary of the first class, 9.5, with a cumulative frequency of 0.

Next, we calculate the cumulative frequencies by adding the frequencies sequentially: 10, 10+18=28, 28+32=60, 60+42=102, 102+28=130, 130+14=144, and 144+6=150. These give us the coordinates to plot: (19.5, 10), (29.5, 28), (39.5, 60), (44.5, 102), (49.5, 130), (54.5, 144), and (59.5, 150).

We then draw the graph with height on the x-axis and cumulative frequency on the y-axis. The points are plotted and joined with a smooth S-shaped curve (ogive), not straight line segments. The curve starts at (9.5, 0) and ends at (59.5, 150).

Key Takeaways

• Cumulative frequency graphs (ogives) are used to represent grouped data and estimate percentiles.
• Upper class boundaries are found by adding 0.5 to the upper limit of each class when data is given to the nearest unit.
• The graph must be drawn as a smooth curve, not a series of straight lines.

Common Mistakes

• Using the upper class limits (e.g., 19, 29) instead of the upper class boundaries (19.5, 29.5).
• Drawing straight line segments between points instead of a smooth curve.
• Forgetting to start the curve at the lower boundary of the first class with a cumulative frequency of 0.
• Incorrectly calculating the cumulative frequencies.

Things to Be Careful About

• Ensure the axes are correctly labelled and scaled. The x-axis should range from 9.5 to 59.5 and the y-axis from 0 to 150.
• The scale must be at least 1 cm = 10 units on both axes to allow accurate plotting and reading.
• The curve should not go above 150 vertically or beyond the upper boundary horizontally.

Walkthrough

The 30th percentile is the value below which 30% of the observations fall. With a total frequency of 150, we calculate 30% of 150:

This means we need to find the height $h$ such that the cumulative frequency is 45. On the cumulative frequency graph, we locate 45 on the vertical axis (cumulative frequency), move horizontally to the right until we intersect the drawn curve, and then move vertically down to read the corresponding value on the horizontal axis (height in cm).

From the graph, this value is approximately 36 cm. Small variations (35–37 cm) are acceptable due to the inherent imprecision in reading values from a hand-drawn graph.

Key Takeaways

• Percentiles can be estimated from cumulative frequency graphs by finding the corresponding cumulative frequency value and reading off the data value.
• Reading from graphs always carries a tolerance; answers within a reasonable range are accepted.

Common Mistakes

• Using the wrong percentage or calculating the position incorrectly (e.g., using 30 instead of 45).
• Drawing lines in the wrong direction (e.g., starting from the x-axis instead of the y-axis for a percentile).
• Not using the graph to read the value, or reading an incorrect value from the axes.

Things to Be Careful About

• Ensure you are reading from the correct axis. For a percentile (a data value), you start on the cumulative frequency axis (y-axis) and read off the height axis (x-axis).
• The graph must be increasing; if it is decreasing, the reading is incorrect.
• Allow for a small range of acceptable answers (e.g., 35–37 cm) when grading graph readings.

Walkthrough

To estimate the mean and standard deviation for grouped data, we assume that all values within a class are equal to the class midpoint. The midpoints are calculated as the average of the lower and upper class boundaries (e.g., for 10–19, boundaries are 9.5 and 19.5, so midpoint is 14.5).

The mean is the weighted average of the midpoints: $\bar{x} = \frac{\sum fx}{\sum f}$. We compute $\sum fx = 10(14.5) + 18(24.5) + \dots + 6(57) = 5840$. Dividing by the total frequency 150 gives the mean $\frac{5840}{150} \approx 38.93$ cm.

For the standard deviation, we use the formula $\text{Var}(X) = \frac{\sum fx^2}{\sum f} - \bar{x}^2$. We first calculate $\sum fx^2 = 10(14.5)^2 + 18(24.5)^2 + \dots + 6(57)^2 = 244285$. Then $\frac{244285}{150} - (38.9333)^2 \approx 1628.567 - 1515.804 = 112.763$. The standard deviation is $\sqrt{112.763} \approx 10.6$ cm.

Key Takeaways

• Grouped data requires the use of class midpoints to estimate central tendency and dispersion.
• The variance formula $\frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2$ is efficient for calculating variance without needing individual data points.
• Always use upper and lower class boundaries (adding/subtracting 0.5) to find midpoints correctly when data is given to the nearest unit.

Common Mistakes

• Using class limits (e.g., 10 and 19) instead of boundaries (9.5 and 19.5) to calculate midpoints. This gives incorrect midpoints like 14.5 vs 14.5 (actually same here, but for 40-44, limits give 42, boundaries give 42; for 10-19, limits give 14.5, boundaries give 14.5. Wait: (10+19)/2 = 14.5. (9.5+19.5)/2 = 14.5. The midpoint is the same. However, using limits is conceptually wrong and can cause errors in other classes or if boundaries are needed for other calculations).
• Forgetting to square the mean in the variance formula.
• Arithmetic errors in calculating $\sum fx$ or $\sum fx^2$.
• Taking the square root of the variance to get the standard deviation (a common step that is sometimes missed).

Things to Be Careful About

• Ensure midpoints are calculated correctly. Although (lower limit + upper limit)/2 often equals (lower boundary + upper boundary)/2, it is safer to use boundaries.
• Keep intermediate values (like the mean) in fractional form or with high precision to avoid rounding errors in the variance calculation.
• The final answer should be given to an appropriate number of significant figures (e.g., 3 s.f.).

Walkthrough

The factory produces chocolates where 25% are wrapped in red foil. When Indigo picks 8 chocolates at random, each pick has a 0.25 chance of being red and the picks are independent. The number of red chocolates $X$ follows a binomial distribution $X \sim B(8, 0.25)$.

We want "no more than 2" red chocolates, which means $X = 0$, $X = 1$, or $X = 2$. These are mutually exclusive events, so their probabilities add.

For each case, use the binomial formula $P(X = x) = \binom{8}{x}(0.25)^x(0.75)^{8-x}$. The coefficient $\binom{8}{x}$ counts the number of ways to arrange $x$ red chocolates among the 8 positions.

• $P(X = 0)$: all 8 are not red, so $(0.75)^8 = 0.10011$.
• $P(X = 1)$: one red among 8 positions has $\binom{8}{1} = 8$ arrangements, so $8 \times 0.25 \times (0.75)^7 = 0.26697$.
• $P(X = 2)$: two reds among 8 positions has $\binom{8}{2} = 28$ arrangements, so $28 \times (0.25)^2 \times (0.75)^6 = 0.31146$.

Adding gives $0.10011 + 0.26697 + 0.31146 = 0.67854$, which rounds to 0.679.

Key Takeaways

The binomial distribution $B(n, p)$ models the number of successes in $n$ independent trials. The formula $P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}$ is the core tool, and "no more than 2" means summing over $X = 0, 1, 2$.

Common Mistakes

• Forgetting the $\binom{8}{x}$ coefficient — this undercounts the number of arrangements of the red chocolates.
• Using $p = 0.75$ instead of $p = 0.25$ — red foil is 25%, not 75%.
• Computing $P(X < 2)$ instead of $P(X \leq 2)$ — omitting the $X = 2$ term.

Things to be Careful About

• The events $X = 0$, $X = 1$, $X = 2$ are mutually exclusive, so their probabilities add.
• The mark scheme accepts an unsimplified correct expression, but the final answer must be rounded to 3 significant figures: 0.679.

Walkthrough

Jake picks chocolates one at a time until he finds a red-foil chocolate. Each pick is red with probability 0.25 and not red with probability 0.75. The first red occurs on the $k$-th choice with probability $(0.75)^{k-1}(0.25)$ — the first $k-1$ picks must all be not red, and the $k$-th must be red.

"Before the 7th choice" means the first red appears at choice 1, 2, 3, 4, 5, or 6. We could sum these six geometric probabilities, but the complement is simpler: the first red is NOT before the 7th choice exactly when the first 6 choices are all not red, with probability $(0.75)^6$.

So $P = 1 - (0.75)^6 = 1 - 0.17798 = 0.82202 \approx 0.822$.

Key Takeaways

The geometric distribution describes the number of trials until the first success. The complement rule $P(A) = 1 - P(A')$ turns a potentially long sum into a single calculation.

Common Mistakes

• Using $(0.75)^7$ instead of $(0.75)^6$ — "before the 7th choice" means the first 6 choices, so we need 6 non-red picks.
• Summing only 5 terms instead of 6 when using the direct geometric sum.

Things to Be Careful About

• "Before the 7th" includes choices 1 through 6.
• The answer must be given to at least 3 significant figures: 0.822.

Walkthrough

We need $P(\text{2nd gold} \mid \text{5th is first unwrapped})$. By the conditional probability formula:

First find $P(B) = P(\text{5th is first unwrapped})$. This means the first 4 chocolates are all wrapped (gold or red, probability 0.55 each) and the 5th is unwrapped (probability 0.45). So:

Next find $P(A \cap B) = P(\text{2nd gold AND 5th is first unwrapped})$. The 5th being the first unwrapped forces the 1st, 2nd, 3rd, 4th to all be wrapped. The 2nd being the first gold forces the 1st to be wrapped but NOT gold — that is, red. So the sequence is:

1st red (0.25), 2nd gold (0.30), 3rd wrapped (0.55), 4th wrapped (0.55), 5th unwrapped (0.45).

Finally:

The exact value is $\frac{30}{121}$.

Key Takeaways

• Conditional probability is computed via $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
• The condition "5th is first unwrapped" forces the first 4 to be wrapped.
• The condition "2nd is first gold" forces the 1st to be red (wrapped but not gold).

Common Mistakes

• Using 0.55 for the 1st chocolate instead of 0.25 — the 1st is red, not merely wrapped.
• Forgetting the denominator $P(\text{5th is first unwrapped})$ — the problem asks for a conditional probability.
• Using $0.45$ alone for the probability the 5th is unwrapped, instead of $(0.55)^4(0.45)$.

Things to Be Careful About

• The sequence must be exactly: red, gold, wrapped, wrapped, unwrapped.
• Keep full precision in intermediate values before dividing.
• The final answer is 0.248 (or $\frac{30}{121}$), to 3 significant figures.

Walkthrough

The word HAPPINESS has 9 letters: H, A, P, P, I, N, E, S, S. Looking at the letters, we can see that the letter P appears twice and the letter S appears twice, while H, A, I, N, E each appear exactly once. So there are two pairs of repeated letters.

When arranging letters with repetitions, the number of distinct arrangements is given by the multinomial coefficient: we divide the total number of arrangements without considering repetitions ($9!$) by the factorial of the count of each repeated letter. We must divide by $2!$ for the two indistinguishable P's and by another $2!$ for the two indistinguishable S's.

So the number of distinct arrangements equals $\frac{9!}{2! \cdot 2!} = \frac{362880}{4} = 90720$.

Key Takeaways

• The formula for arrangements of $n$ letters with repeated counts $a, b, \ldots$ is $\frac{n!}{a! \cdot b! \cdot \ldots}$
• Identifying every repeated letter pair is the essential first step before applying the formula.

Common Mistakes

• Forgetting to divide by $2!$ for each repeated letter, giving $9! = 362880$ (a huge overcount).
• Dividing by only one $2!$, giving $\frac{9!}{2!} = 181440$ (still an overcount).
• Confusing the count of distinct letters with the count of letter types.

Things to Be Careful About

• HAPPINESS contains two P's and two S's, and these are the only repeated letters.
• Each division by $2!$ removes the overcounting caused by swapping two identical letters.

Walkthrough

The phrase "the first and last letters are not the same as each other" is a restriction that is awkward to count directly because every letter at one end is a candidate for the other. Complement counting is the natural tool: count the unrestricted total and subtract the cases we want to exclude.

The unrestricted total comes straight from part (a): 90720 arrangements.

The excluded cases are those where the first letter equals the last letter. Since HAPPINESS has only two letters that appear more than once, namely P and S, the first and last can both be P or both be S; they cannot be both H, both A, etc. because each of those appears only once.

In the first excluded case, fix P at position 1 and P at position 9. The remaining 7 positions must be filled with H, A, I, N, E, S, S. Among these 7 letters only S is repeated, so the number of distinct arrangements is $\frac{7!}{2!} = 2520$.

In the second excluded case, fix S at position 1 and S at position 9. The remaining 7 positions must be filled with H, A, P, P, I, N, E. Only P is repeated, so the number of distinct arrangements is again $\frac{7!}{2!} = 2520$.

The two excluded cases are mutually exclusive (a single arrangement cannot have both P at position 1 and S at position 1, and similarly at position 9), so we add: $2520 + 2520 = 5040$.

Finally, $90720 - 5040 = 85680$.

Key Takeaways

• Complement counting converts a "different" restriction into an "equal" restriction that is easier to count.
• When fixing the same letter at both ends, remember the remaining letters still have repetitions to account for.
• Cases that share the same letter pair (both P, or both S) are mutually exclusive and may be added.

Common Mistakes

• Counting only one of the two cases (only P at both ends, or only S at both ends), giving 2520 or 181440 instead of 85680.
• Forgetting to divide by $2!$ in the inner 7-letter arrangement, leading to $7! = 5040$ per case and so $90720 - 2 \cdot 5040 = 80640$ (wrong).
• Subtracting from the wrong base (e.g. $9!$ instead of $90720$).

Things to Be Careful About

• The first and last being "the same as each other" means the same letter, not necessarily the same physical instance, so both P and both S cases are possible.
• The mark scheme accepts three equivalent methods; this solution uses Method 1 (complement) which is the most direct.

Walkthrough

Two conditions are imposed: the two P's must be adjacent (the PP block) and there must be exactly two letters between the two S's (the S,S block, with two letters filling the gap).

We first lock down the S,S pattern. The two S's must occupy positions $(i, i+3)$ with $i \in \{1, 2, 3, 4, 5, 6\}$, giving 6 possible positions.

For each such S,S position, the PP block must occupy two consecutive positions, and these positions must be either the two positions between the S's (forming the block SPPS) or two of the remaining 5 positions. The number of valid PP placements depends on the S,S position because the remaining 5 positions may be all on one side (giving more consecutive-position options) or split into two groups (giving fewer).

When the S,S pattern is at the end positions $(1,4)$ or $(6,9)$, the remaining 5 positions are all consecutive on one side, allowing 4 PP placements there plus 1 between the S's, for 5 in total. When the S,S pattern is in one of the middle positions $(2,5), (3,6), (4,7), (5,8)$, the remaining 5 positions are split, allowing 3 PP placements there plus 1 between the S's, for 4 in total.

For every valid PP placement, the 5 remaining positions are filled with the 5 distinct letters H, A, I, N, E in $5! = 120$ ways.

Adding everything: $2 \times 5 \times 120 + 4 \times 4 \times 120 = 1200 + 1920 = 3120$.

Key Takeaways

• "Exactly $k$ letters between A and B" means A and B sit at positions $(i, i+k+1)$. Here $k = 2$, so the two S's sit at positions 3 apart.
• The number of valid placements for a separate block (here PP) depends on the geometric configuration of the remaining positions.
• Using $5!$ for the leftover distinct letters is a useful trick once the positions of all repeated and "between" letters are fixed.

Common Mistakes

• Forgetting to include the placement of PP between the S's (which gives the SPPS super-block).
• Miscounting PP placements for some S,S positions, e.g. writing 4 placements for $(1,4)$ instead of 5.
• Treating the PP block as two separate P's and not enforcing their adjacency.
• Forgetting to multiply by $5!$ for the remaining letters.

Things to Be Careful About

• The "5 placements" for the S,S at $(1,4)$ or $(6,9)$ and the "4 placements" for the middle S,S positions are the counts used by the mark scheme; if you split into the SPPS-subcase and the elsewhere-subcase instead, you get $6! + 5! \times 5 \times 4 = 720 + 2400 = 3120$ which is equivalent.
• Two S's at positions $(1,4)$ and at $(6,9)$ are symmetric and both give 5 PP placements.