A-Level Mathematics May/June 2024 Q7(b): Find the number of different arrangements of the 8 digits in which there is a 2 at the…

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  The 8 digits are 2, 2, 4, 4, 4, 5, 6, 7.

  The arrangement must have a 2 at the beginning and a 2 at the end. This fixes the positions of the two 2s:

  $2 \_ \_ \_ \_ \_ \_ 2$

  This leaves the 6 digits {4, 4, 4, 5, 6, 7} to be arranged in the 6 spaces in the middle.

  Step 1: Find the total number of arrangements of the middle 6 digits.

  These 6 digits include three identical 4s. The number of permutations is:

  Total arrangements = $\frac{6!}{3!} = \frac{720}{6} = 120$

  Step 2: Find the number of arrangements where the three 4s are all together.

  To do this, we treat the three 4s as a single block: (444). Now we are arranging 4 items: {(444), 5, 6, 7}.

  The number of ways to arrange these 4 distinct items is:

  Arrangements with 4s together = $4! = 24$

  (Note: We do not multiply by 3! for the internal arrangement of the (444) block as the 4s are identical).

  Step 3: Find the number of arrangements where the three 4s are not all together.

  This is found by subtracting the 'unwanted' arrangements (where the 4s are together) from the total number of arrangements.

  Number of required arrangements = (Total arrangements) - (Arrangements with 4s together)

  Number of required arrangements = $120 - 24 = 96$

  Final Answer: 96

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  How the marks are awarded

  • M1 — The first M1 is for correctly calculating the total number of arrangements of the 6 middle digits, accounting for the three repeated 4s. This is shown by the calculation $6!/3!$.
  • B1 — The B1 mark is for correctly calculating the number of arrangements where the three 4s are together. This involves treating the three 4s as a single block and arranging the resulting 4 items, giving $4!$.
  • M1 — The second M1 is for the correct method of subtracting the number of arrangements where the 4s are together from the total, as shown by the expression $120 - 24$ or equivalent.
  • A1 — The A1 mark is for the correct final answer of 96.

  Common mistakes

  • Forgetting to divide by 3! for the repeated 4s when finding the total number of arrangements. This leads to calculating $6! = 720$ as the total, resulting in an incorrect final answer of $720 - 24 = 696$.
  • Calculating the number of arrangements with the 4s together as $4! \times 3!$. This mistake arises from incorrectly trying to account for internal arrangements of the three 4s, forgetting that they are identical.

  Examiner tip: For permutation problems involving a 'not together' condition, it is almost always easier to calculate the total number of arrangements and subtract the number of arrangements where the items are together.

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