A-Level Mathematics May/June 2024 Q7(c): Three digits are selected at random from the eight digits 1, 2, 2, 3, 4, 4, 4, 5. Find…

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  To find the probability that the three digits selected are all different, we first need to find the number of ways to select three different digits (favourable outcomes) and divide it by the total number of ways to select any three digits (total outcomes).

  The set of eight digits is {1, 2, 2, 3, 4, 4, 4, 5}. The distinct digits available are {1, 2, 3, 4, 5}.

  Numerator: Number of ways to select 3 different digits

  We must consider all combinations of 3 distinct digits chosen from {1, 2, 3, 4, 5} and account for the repetitions in the original set.

  • Digits {1, 2, 3}: $^1C_1 \times ^2C_1 \times ^1C_1 = 1 \times 2 \times 1 = 2$ ways
  • Digits {1, 2, 4}: $^1C_1 \times ^2C_1 \times ^3C_1 = 1 \times 2 \times 3 = 6$ ways
  • Digits {1, 2, 5}: $^1C_1 \times ^2C_1 \times ^1C_1 = 1 \times 2 \times 1 = 2$ ways
  • Digits {1, 3, 4}: $^1C_1 \times ^1C_1 \times ^3C_1 = 1 \times 1 \times 3 = 3$ ways
  • Digits {1, 3, 5}: $^1C_1 \times ^1C_1 \times ^1C_1 = 1 \times 1 \times 1 = 1$ way
  • Digits {1, 4, 5}: $^1C_1 \times ^3C_1 \times ^1C_1 = 1 \times 3 \times 1 = 3$ ways
  • Digits {2, 3, 4}: $^2C_1 \times ^1C_1 \times ^3C_1 = 2 \times 1 \times 3 = 6$ ways
  • Digits {2, 3, 5}: $^2C_1 \times ^1C_1 \times ^1C_1 = 2 \times 1 \times 1 = 2$ ways
  • Digits {2, 4, 5}: $^2C_1 \times ^3C_1 \times ^1C_1 = 2 \times 3 \times 1 = 6$ ways
  • Digits {3, 4, 5}: $^1C_1 \times ^3C_1 \times ^1C_1 = 1 \times 3 \times 1 = 3$ ways

  Total number of ways to select 3 different digits = $2 + 6 + 2 + 3 + 1 + 3 + 6 + 2 + 6 + 3 = 34$.

  Denominator: Total number of ways to select 3 digits

  We are selecting 3 digits from a total of 8 digits, without regard to order. Total ways = $^8C_3 = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

  Probability

  $P(\text{all 3 digits are different}) = \frac{\text{Number of ways to select 3 different digits}}{\text{Total number of ways to select 3 digits}}$

  $P(\text{all 3 digits are different}) = \frac{34}{56} = \frac{17}{28}$

  As a decimal, this is approximately 0.607 (3 s.f.).

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  How the marks are awarded

  • M1 — Calculating the number of ways for at least one specific combination of 3 distinct digits, for example, showing that for digits {1, 2, 3}, the number of ways is $^1C_1 \times ^2C_1 \times ^1C_1 = 2$.
  • A1 — Correctly calculating the number of ways for at least five of the ten possible combinations of 3 distinct digits.
  • M1 — Demonstrating a complete method for the numerator by summing the ways for all ten combinations of distinct digits, leading to the total of 34.
  • B1 — Correctly identifying or using the total number of ways to select 3 digits from 8 as the denominator, calculated as $^8C_3 = 56$.
  • A1 — Stating the final correct probability as $\frac{34}{56}$ or the simplified fraction $\frac{17}{28}$, or a correct decimal equivalent like 0.607.

  Common mistakes

  • Incorrectly calculating the total number of outcomes (denominator). A common error is to use the 5 distinct digit types, e.g., $^5C_3$, instead of the 8 available digits, $^8C_3$.
  • Ignoring the repeated digits when calculating the numerator. For example, simply stating there are $^5C_3 = 10$ ways to choose 3 different digits, which fails to account for the multiple '2's and '4's available for selection.
  • Failing to be systematic and missing one or more of the ten possible cases for the numerator. For instance, only calculating cases involving the digit '1' and forgetting combinations like {2, 3, 4}.
  • Using the indirect method (1 - P(not all different)) and making errors. For example, when calculating ways to get two '4's and one other digit, a mistake would be $^3C_2 \times ^4C_1$ (choosing from 4 digit types) instead of the correct $^3C_2 \times ^5C_1$ (choosing from the 5 other digits).

  Examiner tip: When selecting from a set with repeated items, you must break the problem down into distinct cases based on the composition of your selection and then sum the number of ways for each case.

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