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"In how many arrangements of the letters do all the vowels come together?" is one of the most common counting questions in school and JEE papers. The method is short, but students lose marks by forgetting to unpack the bundle, by mishandling repeated letters, or by confusing "not all together" with "no two together". This post sets out the bundle method in three steps and works through five examples, each checked by brute-force counting.
Where is this in the syllabus?
NCERT's Class 11 Permutations and Combinations chapter (2026-27 reprint) introduces the idea in Example 14, with the word DAUGHTER: it treats the three vowels "as a single object" and multiplies by the 3! ways of arranging them inside. The JEE (Main) 2026 syllabus covers this under permutations and "simple applications". We use different words below so you can test yourself on NCERT's examples separately. Check your exam year's syllabus.
What are the three steps?
Bundle. Tie the letters that must be together into one block. Treat the block as a single item.
Arrange. Count the arrangements of the block plus the remaining items.
Unpack. Multiply by the number of ways to arrange the letters inside the block.
If letters repeat, divide by the factorials of repeats, but do it separately: repeats inside the block in step 3, repeats outside in step 2.
Worked example 1: EQUATION
EQUATION has 8 different letters: five vowels (E, U, A, I, O) and three consonants (Q, T, N).
All vowels together. Bundle the five vowels. Items to arrange: the block plus Q, T, N, which is 4 items, so 4! = 24. Unpack: 5! = 120 orders inside. Answer: 24 × 120 = 2880.
Vowels not all together. Total arrangements are 8! = 40,320. Subtract: 40,320 − 2880 = 37,440.
Vowels together and consonants together. Now there are two blocks, which can be arranged in 2! ways. Unpack both: 5! × 3!. Answer: 2 × 120 × 6 = 1440.
A brute-force count by computer over all 40,320 arrangements gives the same three numbers.
Worked example 2: repeated letters inside the bundle
BALLOON has 7 letters: B, A, L, L, O, O, N. The vowels are A, O, O. How many arrangements have the vowels together?
Bundle A, O, O. Outside items: the block, B, L, L, N. That is 5 items with L repeated twice: 5!/2! = 60.
Unpack the block: A, O, O can be arranged in 3!/2! = 3 ways (AOO, OAO, OOA).
Answer: 60 × 3 = 180.
The total number of arrangements of BALLOON is 7!/(2! × 2!) = 1260, so 1260 − 180 = 1080 have the vowels not all together.
Worked example 3: a bundle of identical letters
MISSISSIPPI has 11 letters: M once, I four times, S four times, P twice. Total arrangements: 11!/(4! × 4! × 2!) = 34,650.
All four S's together. Bundle SSSS. Items: the block, M, four I's and two P's, which is 8 items. Arrange: 8!/(4! × 2!) = 840. Unpack: the four S's are identical, so there is only 1 way inside. Answer: 840.
All S's together and all I's together. Now two blocks, plus M and two P's: 5 items with P twice, so 5!/2! = 60.
Here the unpacking step contributes 1, which is why students who forget it sometimes get this question right by luck and the EQUATION one wrong.
Worked example 4: together and apart in one question
Six people A, B, C, D, E and F stand in a row. A and B must stand together, but C and D must not. How many arrangements?
A and B together: bundle them. 5 items give 5! = 120, and the bundle unpacks as AB or BA, so 240.
Of these, count the ones where C and D are also together: two bundles plus E and F gives 4 items, 4! = 24, times 2 × 2 for the two bundles, so 96.
Answer: 240 − 96 = 144.
The pattern is "bundle what must be together, then subtract the cases where the forbidden pair is also bundled".
How is "no two together" different?
Take ORANGE, with vowels O, A, E and consonants R, N, G. There are 6! = 720 arrangements.
All vowels together: 4! × 3! = 144.
Vowels not all together: 720 − 144 = 576. This still includes arrangements like ROANGE, where two vowels touch.
No two vowels together: use the gap method instead. Arrange the consonants in 3! = 6 ways, which creates 4 gaps (_R_N_G_). Place the three vowels in three different gaps in P(4, 3) = 24 ways. Answer: 6 × 24 = 144.
That both answers are 144 is a coincidence of this word, not a rule. "Not all together" and "no two together" are different questions with different methods. Underline which one is asked before you start.
Which mistakes cost the most marks?
Forgetting to unpack. Answering 4! for EQUATION instead of 4! × 5!.
Counting the bundle as several items. The block is one item in step 2.
Dividing a repeat twice, or dividing an outside repeat when unpacking.
Using "total minus together" for "no two together". Use gaps.
Treating a row as a circle. For round tables, see our post on circular permutations.
A practice routine
When you check a solution, explain each step to yourself: why this is one item, why we divide here. In a study of students learning from worked examples, Chi and colleagues (1989) found that "Good" students generated many self-explanations that related each step to principles, while "Poor" students produced few and relied heavily on the examples.
Day 1: Five "all together" questions on words with no repeats. Write the three steps every time.
Day 2: Five words with repeats, such as COMMITTEE, BANANA-style words and your own name. Mark each repeat as inside or outside the bundle.
Day 3: Five mixed questions with "not all together" and "no two together" side by side.
Day 4: Shrink one question to four letters and list every arrangement by hand.
If you get stuck on why a step works, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Try this now: in how many arrangements of COMMITTEE do the two M's, two T's and two E's each stay together as pairs?
