Permutations and Combinations: Numerical Problems Set (1)

easy 2 min read

Question

How many 4-letter words (with or without meaning) can be formed using the letters of the word “EQUATION” if no letter is repeated?

Solution — Step by Step

EQUATION has 8 distinct letters: E, Q, U, A, T, I, O, N.

We need to choose 4 letters from 8 and arrange them — that’s 8P4^8P_4.

8P4=8!(84)!=8!4!=8×7×6×5=1680^8P_4 = \frac{8!}{(8-4)!} = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680

Final answer: 16801680 words.

Why This Works

A “word” here means an ordered arrangement of letters. So this is a permutation, not a combination. The first slot can be filled by any of 8 letters, the second by any of the remaining 7, and so on — the multiplication principle gives 8×7×6×5=16808 \times 7 \times 6 \times 5 = 1680.

If the question had asked about choosing 4 letters (set selection, no order), it would be (84)=70\binom{8}{4} = 70.

Alternative Method

Fill slots one by one:

  • Slot 1: 8 choices
  • Slot 2: 7 choices (one letter used)
  • Slot 3: 6 choices
  • Slot 4: 5 choices
  • Total: 8×7×6×5=16808 \times 7 \times 6 \times 5 = 1680.

This is the multiplication principle in raw form, identical to the formula.

Permutation = “arrangement” (order matters). Combination = “selection” (order doesn’t matter). When the question uses words like “arrangements,” “words,” “queue,” or “order,” it’s a permutation. When it uses “team,” “committee,” or “selection,” it’s a combination.

Common Mistake

Confusing “no letter repeated” with “use each letter once” — they’re the same here because we have 8 distinct letters. But if the source word had repeated letters (like “BANANA”), we’d have to deal with identical letters, which is a different formula.

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