Year 11: How to solve permutation with repetition word problems in probability

Back to tutorials

In Year 11 Mathematics in Australia, you often use permutations to count how many different arrangements are possible. Sometimes, objects repeat. When that happens, we use permutations with repetition. This article shows you, step by step, how to solve a probability word problem that uses permutations with repetition.

1. What is a permutation with repetition?

A permutation is an arrangement of objects where the order matters.

If all objects are different, the number of permutations of n objects is:

n! = n × (n − 1) × (n − 2) × … × 2 × 1

If some objects are the same (repeated), we use the formula:

Number of different permutations =
N! / (n₁! × n₂! × n₃! × …)

• N is the total number of objects.
• n₁, n₂, n₃, … are the numbers of each repeated object.

2. Link to probability

In probability, we often assume that each different arrangement is equally likely. For example, if letters are shuffled at random, all distinct permutations are equally likely.

To find a probability:

• First find the number of total outcomes (all possible different arrangements).
• Then find the number of favourable outcomes (arrangements that satisfy the condition).
• Use the rule: Probability = favourable / total.

3. A typical word problem

4. Step-by-step method: “How to solve it”

Step 1: Understand and organise the information

Write the word clearly and list the frequency of each letter:

STATISTICS

• S appears 3 times
• T appears 3 times
• A appears 1 time
• I appears 2 times
• C appears 1 time

Check the total number of letters:

3 (S) + 3 (T) + 1 (A) + 2 (I) + 1 (C) = 10 letters

Step 2: Identify what you are asked to find

• (a) Total number of different permutations of all 10 letters.
• (b) The probability that the first and last letters are both S.

Step 3: Choose the correct counting method

Because some letters repeat, we must use the:

Permutation with repetition formula:
N! / (n₁! × n₂! × …)

For part (b), we will:

1. Count the number of arrangements where the first and last letters are S.
2. Divide by the total number from part (a).

Step 4: Solve part (a) carefully

Total letters: N = 10

Repeated letters:

• S: 3 times
• T: 3 times
• I: 2 times
• A: 1 time
• C: 1 time

Number of distinct permutations:

Total arrangements = 10! / (3! × 3! × 2! × 1! × 1!)

You do not have to multiply by 1! because 1! = 1, but it can help you see the pattern.

Now simplify:

• 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3 628 800
• 3! = 6, so 3! × 3! = 6 × 6 = 36
• 2! = 2

So the denominator is:

3! × 3! × 2! = 36 × 2 = 72

Total arrangements:

3 628 800 / 72 = 50 400

So, there are 50 400 different arrangements of the letters in STATISTICS.

Step 5: Solve part (b) – favourable outcomes

We want arrangements that:

• start with S, and
• end with S.

Fix an S at the beginning and an S at the end:

S _ _ _ _ _ _ _ _ S

We have used 2 of the 3 S letters, so there is:

• 1 S remaining,
• 3 T letters,
• 2 I letters,
• 1 A,
• 1 C.

We now have 8 positions in the middle and 8 letters left in total:

Total middle letters = 1 (S) + 3 (T) + 2 (I) + 1 (A) + 1 (C) = 8

We again use the permutation with repetition formula for these 8 letters:

Number of favourable arrangements = 8! / (3! × 2! × 1! × 1! × 1!)

Simplify:

• 8! = 40 320
• 3! = 6
• 2! = 2

Denominator:

3! × 2! = 6 × 2 = 12

Favourable arrangements:

40 320 / 12 = 3 360

So there are 3 360 arrangements that start and end with S.

Step 6: Compute the probability

We now use:

Probability = (number of favourable outcomes) / (total number of outcomes)

So:

P(first and last letters are S) = 3 360 / 50 400

Now simplify the fraction. Divide numerator and denominator by 120:

• 3 360 / 120 = 28
• 50 400 / 120 = 420

So:

3 360 / 50 400 = 28 / 420

Next, divide numerator and denominator by 28:

• 28 / 28 = 1
• 420 / 28 = 15

Therefore:

P(first and last letters are S) = 1 / 15

So the probability that a random arrangement of the letters in STATISTICS starts and ends with S is 1/15.

5. Another example (shorter practice)

Step A: Count total arrangements.

BALLOON has:

• B: 1
• A: 1
• L: 2
• O: 2
• N: 1

Total letters: N = 7

Total number of different arrangements:

7! / (2! × 2!) = 5 040 / 2 = 2 520 (since 7! = 5 040 and 2! × 2! = 4)

Step B: Count favourable arrangements (L letters together).

Treat the two L letters as a single block, call it [LL].

Now objects: [LL], B, A, O, O, N

• Total objects: 6
• O repeats 2 times.

Number of permutations:

6! / 2! = 720 / 2 = 360

Step C: Probability.

P(L letters together) = 360 / 2 520

Simplify by dividing numerator and denominator by 120:

• 360 / 120 = 3
• 2 520 / 120 = 21

Then divide both by 3:

• 3 / 3 = 1
• 21 / 3 = 7

So:

P(L letters together) = 1 / 7

6. General “how to solve it” checklist

1. Read the problem slowly. Underline key words like “arranged”, “in how many ways”, “probability”, “begins with”, “together”, etc.
2. List the objects and their frequencies. Count how many times each symbol or letter appears.
3. Find the total number of arrangements. Use N! / (n₁! × n₂! × …).
4. Count favourable outcomes.
   • Fix certain positions (e.g. start with S, end with S).
   • Group objects when needed (e.g. [LL] as one block).
   • Apply the same permutation-with-repetition formula to what is left.
5. Compute the probability. Use favourable / total and simplify the fraction.
6. Check if your answer makes sense. The probability must be between 0 and 1.

7. Common mistakes and tips

8. Final summary

Permutations with repetition are essential when objects repeat, and they appear often in word problems in probability. The key idea is:

Count carefully using the formula N! / (n₁! × n₂! × …) for both the total number of arrangements and the number of favourable arrangements.

Then, probability is just:

Probability = favourable outcomes / total outcomes

If you follow the step-by-step plan and watch out for the common mistakes, you will be in a strong position to tackle Year 11 questions on permutations with repetition in probability.