A-Level Mathematics May/June 2024 Q7(b): How many different arrangements are there of the 10 letters in the word REGENERATE in w…

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  The word is REGENERATE, which has 10 letters: R, R, E, E, E, E, G, N, A, T.

  Constraint 1: The 4 Es are together. We treat the four Es as a single block: (EEEE). Let's call this block $E_B$.

  Constraint 2: The 2 Rs have exactly 3 letters in between them. This forms another block structure: R _ _ _ R. Let's call this block $R_B$.

  The distinct single letters available are G, N, A, T.

  Step 1: Form the R _ _ _ R block. The three letters between the two Rs must be chosen from the four distinct single letters {G, N, A, T}. The block of Es, $E_B$, cannot be placed here as it is not a single letter.

  The number of ways to choose 3 letters from 4 and arrange them in the spaces is a permutation: Number of ways to fill the gaps = $^4P_3 = \frac{4!}{(4-3)!} = 4! = 24$.

  Step 2: Arrange the blocks. After forming the block $R_B$, we have used the two Rs and three of the single letters. The items we now need to arrange are:

  1. The block $R_B$ (e.g., R G N A R).
  2. The block $E_B$ (i.e., EEEE).
  3. The one remaining single letter from {G, N, A, T} that was not used in $R_B$.

  These are 3 distinct items. The number of ways to arrange these 3 items is: Number of ways to arrange the blocks = $3! = 6$.

  Step 3: Calculate the total number of arrangements. The total number of arrangements is the product of the number of ways from Step 1 and Step 2. Total arrangements = (Ways to form $R_B$) $\times$ (Ways to arrange the items) Total arrangements = $4! \times 3!$ Total arrangements = $24 \times 6$ Total arrangements = $144$.

  There are 144 different arrangements.

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  How the marks are awarded

  • M1 — For identifying that the 3 letters between the Rs must be chosen and arranged from the 4 available distinct letters {G, N, A, T}. This is calculated as a permutation $^4P_3$, which is equal to $4!$. This is seen or implied by the value 24.
  • M1 — For realising that after forming the 'R...R' block, there are three items to arrange: the 'R...R' block itself, the '(EEEE)' block, and the one leftover single letter. The number of ways to arrange these three items is $3!$ (or 6).
  • A1 — For the correct, unsimplified expression resulting from multiplying the outcomes of the two independent steps, which is $4! \times 3!$.
  • B1 — For the final correct answer of 144, with correct supporting working shown (WWW - working must be watched).

  Common mistakes

  • Using combinations instead of permutations for the letters between the Rs ($^4C_3$ instead of $^4P_3$), which ignores the internal arrangement of those three letters.
  • Forgetting to arrange the blocks. A student might correctly calculate $^4P_3 = 24$ but fail to multiply by $3!$ for the arrangement of the resulting super-items.
  • Incorrectly identifying the number of items to arrange in the final step. For example, after creating the R...R block, thinking there are 5 items left to arrange (the E-block and all 4 of G,N,A,T), forgetting 3 are already used.
  • Attempting to place the block of four Es between the two Rs. The question specifies '3 letters', which implies single characters, not a block of four.

  Examiner tip: When dealing with multiple constraints in permutation problems, group the constrained items into blocks first and then treat those blocks as single items in the final arrangement.

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