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Worked Examples · Example 16

Q.Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,

(i) do the words start with P
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with I and end in P?
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INDEPENDENCE has 12 letters with repetitions: I(1), N(3), D(2), E(4), P(1), C(1). Total arrangements: 12!3!⋅2!⋅4!\frac{12!}{3! \cdot 2! \cdot 4!}. Then apply constraints for each part.

The word INDEPENDENCE contains 12 letters, but several appear more than once. When letters repeat, we must account for indistinguishable arrangements—swapping two identical N's doesn't create a new word.

First, let's count the frequency of each letter:

  • I: 1
  • N: 3
  • D: 2
  • E: 4
  • P: 1
  • C: 1

The fundamental principle here is permutations with repetition: if we have nn total objects where certain types repeat (say n1n_1 of type 1, n2n_2 of type 2, etc.), the number of distinct arrangements is:

n!n1!⋅n2!⋅…⋅nk!\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}

The factorials in the denominator "cancel out" the overcounting from identical objects.

Total arrangements of INDEPENDENCE

  1. We have 12 positions to fill with our letters.
  2. If all letters were distinct, we'd have 12!12! arrangements.
  3. But N appears 3 times (overcounting by 3!3!), D appears 2 times (overcounting by 2!2!), and E appears 4 times (overcounting by 4!4!).
  4. The number of distinct arrangements is:

12!3!⋅2!⋅4!=4790016006⋅2⋅24=479001600288=1663200\frac{12!}{3! \cdot 2! \cdot 4!} = \frac{479001600}{6 \cdot 2 \cdot 24} = \frac{479001600}{288} = 1663200

(i) Arrangements starting with P

  1. Fix P in the first position.
  2. We now arrange the remaining 11 letters: I(1), N(3), D(2), E(4), C(1).
  3. These 11 letters go into 11 positions:

11!3!⋅2!⋅4!=399168006⋅2⋅24=39916800288=138600\frac{11!}{3! \cdot 2! \cdot 4!} = \frac{39916800}{6 \cdot 2 \cdot 24} = \frac{39916800}{288} = 138600

(ii) All vowels occur together

The vowels in INDEPENDENCE are: I, E, E, E, E (5 vowels total).

  1. Treat the vowel block as a single unit. Now we have 8 objects to arrange:

    • The vowel-block (1 unit)
    • N, N, N (3)
    • D, D (2)
    • P (1)
    • C (1)

  2. Arrange these 8 objects:

8!3!⋅2!=403206⋅2=4032012=3360\frac{8!}{3! \cdot 2!} = \frac{40320}{6 \cdot 2} = \frac{40320}{12} = 3360

  1. Within the vowel block, arrange I, E, E, E, E (5 letters with E repeated 4 times):

5!4!=12024=5\frac{5!}{4!} = \frac{120}{24} = 5

  1. Total arrangements with vowels together:

3360×5=168003360 \times 5 = 16800

(iii) Vowels never occur together

This is the complement of "vowels always together."

  1. Total arrangements (from above): 16632001663200
  2. Arrangements where vowels are together (from part ii): 1680016800
  3. Arrangements where vowels are never all together: …
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