How Many Words Can Be Formed From the Letters of the Word DAUGHTER Such That All Vowels Are Not Together

JEE Main 2025 — 24 January, Evening Shift. Previous Year Question.

Problem

How many words can be formed from the letters of the word DAUGHTER such that all vowels are not together?

Key insight. “Not all together” is the complement of “all together” — so rather than directly counting the messier condition, it’s far simpler to count all arrangements, subtract the ones where the vowels are all bunched into one block, and let the subtraction do the work.

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Approach

Count the total number of arrangements of all 8 distinct letters of DAUGHTER with no restriction. Separately, count the arrangements where all 3 vowels (A, U, E) are forced together by treating them as a single block alongside the 5 consonants, then multiplying by the ways to arrange the vowels within that block. Subtract the second count from the first.

Solution

Step 1 — Identify the letters

DAUGHTER has 8 distinct letters: D, A, U, G, H, T, E, R — of which A, U, E are vowels (3 vowels) and D, G, H, T, R are consonants (5 consonants).

Step 2 — Count all arrangements with no restriction

$$8! = 40320$$

Step 3 — Count arrangements where all vowels are together

Treat the 3 vowels (A, U, E) as a single block. Together with the 5 consonants, this gives $5+1=6$ units to arrange:

$$6! \text{ ways to arrange the 6 units}$$

Within the vowel block, the 3 vowels can be arranged among themselves in:

$$3! \text{ ways}$$

Total arrangements with all vowels together:

$$6!\times3! = 720\times6 = 4320$$

Step 4 — Subtract to find “not all together”

$$8! – 6!\times3! = 40320-4320 = 36000$$

Answer

Common mistakes

• Trying to directly count “vowels not together” without using the complement. Directly enumerating every arrangement where vowels are scattered (not all adjacent) is far more complex than counting the “all together” case and subtracting — the complement trick is what makes this problem tractable.
• Forgetting to multiply by $3!$ for the internal arrangement of the vowel block. Treating the 3 vowels as a single unit accounts for their position relative to the consonants, but the vowels themselves (A, U, E) are still distinct and can be ordered among themselves in $3!$ ways within that block.

Practise next

• How many words can be formed from the letters of the word EQUATION such that all vowels are not together? (Note: EQUATION has 5 vowels and 3 consonants — use the same total-minus-together approach.)