Grade 12 Maths · Probability · Count arrangements with repeated items

Count arrangements with repeated items Grade 12

Count arrangements when some items are identical by dividing by the factorial of each repeat, for example the letters of BOOK can be arranged in 4! 2! = 12 ways.

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Practice · NavyBlue · 2 marks

How many different arrangements can be made with all the letters of the word BANANA?(2)

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3. Practice · Siyavula Grade 12 Maths · Ex 10.7 · 3 marks
You have the word 'ASSESSMENT'.
(c) If the first and last letters are identical, how many letter arrangements are there?(3)
Show the memo (3 ticks)Memo
1. ✓
  S at both endsRemaining: S, S, E, E, A, M, N, T: $\frac{8 !}{2 ! \times 2 !} = 10080$
  1 mark · Count arrangements with repeated items
2. ✓
  E at both endsRemaining: S, S, S, S, A, M, N, T: $\frac{8 !}{4 !} = 1680$
  1 mark · Count arrangements with repeated items
3. ✓
  Total $10080 + 1680 = 11760$
  1 mark · Count arrangements with a condition
Question adapted from Siyavula Education (CC BY 3.0) · Solution by NavyBlue
4. Practice · Siyavula Grade 12 Maths · Ex 10.7 · 3 marks
You have the word 'EXCELLENT'.
(c) If the first and last letters are identical, how many letter arrangements are there?(3)
Show the memo (3 ticks)Memo
1. ✓
  E at both endsRemaining: E, X, C, L, L, N, T: $\frac{7 !}{2 !} = 2520$
  1 mark · Count arrangements with repeated items
2. ✓
  L at both endsRemaining: E, E, E, X, C, N, T: $\frac{7 !}{3 !} = 840$
  1 mark · Count arrangements with repeated items
3. ✓
  Total $2520 + 840 = 3360$
  1 mark · Count arrangements with a condition
Question adapted from Siyavula Education (CC BY 3.0) · Solution by NavyBlue
5. Practice · Siyavula Grade 12 Maths · Ex 10.8 · 4 marks
If the letters of the word 'COMMITTEE' are randomly arranged, what is the probability that the letter arrangements start and end with the same letter?(4)
Show the memo (4 ticks)Memo
1. ✓
  Total arrangements $\frac{9 !}{2 ! \times 2 ! \times 2 !} = 45360$
  1 mark · Count arrangements with repeated items
2. ✓
  Same letter at both endsWith M at both ends: $\frac{7 !}{2 ! \times 2 !} = 1260$ ; the same for T and for E
  1 mark · Count arrangements with repeated items
3. ✓
  Favourable $3 \times 1260 = 3780$
  1 mark · Count arrangements with a condition
4. ✓
  Probability $P = \frac{3 780}{45 360} = \frac{1}{12}$
  1 mark · Calculate a simple probability
Question adapted from Siyavula Education (CC BY 3.0) · Solution by NavyBlue
6. Practice · NavyBlue · 2 marks
Consider the letters of the word LETTER.
In how many of these arrangements are the two Ts NOT next to each other?(2)
Show the memo (2 ticks)Memo
1. ✓
  Ts togetherTT as one block: $\frac{5 !}{2 !} = 60$
  1 mark · Count arrangements with repeated items
2. ✓
  Answer $180 - 60 = 120$
  1 mark · Count using the "not" rule

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