Mathematics 9709/62 — October/November 2018

Walkthrough

The word MISSISSIPPI has 11 letters, but many of them are repeated. If all 11 letters were different, there would be $11!$ arrangements. However, the four I's are identical, the four S's are identical, and the two P's are identical. Swapping identical letters does not produce a different arrangement, so we divide by $4!$ for the I's, $4!$ for the S's, and $2!$ for the P's.

This gives the number of distinct arrangements as:

$$\frac{11!}{4!4!2!} = 34650$$

Key Takeaways

When arranging objects where some are identical, use the multinomial coefficient $\frac{n!}{a!b!c!\cdots}$, where $a, b, c, \ldots$ are the frequencies of the repeated items. This removes the overcounting caused by identical letters.

Common Mistakes

• Using $11!$ without dividing by the factorials of repeated letters.
• Miscounting the letters: M appears once, I appears four times, S appears four times, P appears twice.
• Dividing by the wrong factorials.
• Not simplifying the final answer.

Things to Be Careful About

The mark scheme awards the method mark for the correct factorial division and the accuracy mark for the final answer 34650. Always show the formula $\frac{11!}{4!4!2!}$ to make the method clear.

Walkthrough

Choosing two letters at random from 11 means choosing an unordered pair. The total number of possible pairs is $\binom{11}{2} = 55$.

For the two letters to be the same, both letters must come from the same repeated-letter group:

• There are 4 S's, so the number of ways to choose two S's is $\binom{4}{2} = 6$.
• There are 4 I's, so the number of ways to choose two I's is $\binom{4}{2} = 6$.
• There are 2 P's, so the number of ways to choose two P's is $\binom{2}{2} = 1$.
• There is only 1 M, so it is impossible to choose two M's.

These cases are mutually exclusive, so add them:

$$6 + 6 + 1 = 13$$

Therefore the probability is:

$$\frac{13}{55}$$

An equivalent sequential method is:

$$P(SS) + P(II) + P(PP) = \frac{4}{11}\cdot\frac{3}{10} + \frac{4}{11}\cdot\frac{3}{10} + \frac{2}{11}\cdot\frac{1}{10} = \frac{26}{110} = \frac{13}{55}$$

Key Takeaways

When selecting unordered pairs from groups with identical items, use combinations to count favourable outcomes. Probability is the number of favourable outcomes divided by the total number of outcomes. When several disjoint cases give the required outcome, add their probabilities or counts.

Common Mistakes

• Forgetting that P appears twice, so a pair of P's is possible.
• Including M as a possible same-letter pair, since there is only one M.
• Counting ordered pairs in the numerator while using an unordered denominator, or vice versa.
• Not simplifying $\frac{26}{110}$ to $\frac{13}{55}$.
• Using permutations instead of combinations when the order of choosing the two letters does not matter.

Things to Be Careful About

The mark scheme allows either a counting method or a sequential probability method. For the counting method, use $\binom{11}{2}$ as the denominator and $\binom{4}{2} + \binom{4}{2} + \binom{2}{2}$ as the numerator. For the sequential method, multiply probabilities for each pair and add the three mutually exclusive cases. Avoid mixing ordered and unordered counts.