A-Level Mathematics October/November 2024 Q7(c): Four letters are selected at random from the 9 letters in the word INTELLECT. (c) Find…

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  The letters in the word INTELLECT are I, N, T, T, E, E, L, L, C. Total number of letters is 9. We are selecting 4 letters.

  The condition is that the selection contains 'at least one E' and 'exactly one T'. This can be broken down into two mutually exclusive cases:

  Case 1: Exactly one T and exactly one E. We select 1 T from the 2 available T's: $^2C_1$ We select 1 E from the 2 available E's: $^2C_1$ We must select the remaining 2 letters from the other letters which are not T or E. These are {I, N, C, L, L}, a total of 5 letters. Number of ways to choose 2 from these 5 letters: $^5C_2$ Total for Case 1 = $^2C_1 \times ^2C_1 \times ^5C_2 = 2 \times 2 \times \frac{5 \times 4}{2} = 4 \times 10 = 40$.

  Case 2: Exactly one T and two E's. We select 1 T from the 2 available T's: $^2C_1$ We select 2 E's from the 2 available E's: $^2C_2$ We must select the remaining 1 letter from the other 5 letters {I, N, C, L, L}. Number of ways to choose 1 from these 5 letters: $^5C_1$ Total for Case 2 = $^2C_1 \times ^2C_2 \times ^5C_1 = 2 \times 1 \times 5 = 10$.

  Total number of favourable selections = Sum of cases = $40 + 10 = 50$.

  The total number of possible selections of 4 letters from 9 is $^9C_4$. $^9C_4 = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.

  The required percentage is: Percentage = $\frac{\text{Favourable selections}}{\text{Total selections}} \times 100\%$ Percentage = $\frac{50}{126} \times 100\% = 39.6825...\%$

  Percentage $\approx 39.7\%$ (to 3 s.f.)

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  How the marks are awarded

  • B1 — Correctly calculating the number of ways for the case with one T and one E. This is shown in the model answer as '$^2C_1 \times ^2C_1 \times ^5C_2 = 40$'.
  • B1 — Correctly calculating the number of ways for the case with one T and two E's. This is shown in the model answer as '$^2C_1 \times ^2C_2 \times ^5C_1 = 10$'.
  • M1 — Summing the results of the valid cases and dividing by the total number of possible selections, $^9C_4$. The working '$\frac{40+10}{126}$' or '$\frac{50}{^9C_4}$' earns this mark.
  • A1 — Obtaining the final correct answer of 39.7% or any value in the range [39.68, 39.7].

  Common mistakes

  • Forgetting to consider the case with two E's, and only calculating the result for exactly one E, leading to an answer of 40/126.
  • Incorrectly identifying the pool of 'other' letters to choose from. For example, after picking one T and one E, choosing the remaining two letters from the 7 non-T letters (which incorrectly includes an E) instead of the 5 letters that are not T or E.
  • Calculating the total number of selections incorrectly, for instance by using permutations ($^9P_4$) instead of combinations ($^9C_4$).
  • Leaving the answer as a fraction (50/126 or 25/63) or a decimal (0.397) instead of a percentage as required by the question.

  Examiner tip: When a selection problem involves conditions like 'at least one', systematically break it down into mutually exclusive cases (e.g., exactly one, exactly two, etc.) and sum the results.

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