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"No two girls sit together." "No two vowels are adjacent." "Choose three seats with no two next to each other." These look like three topics, but one method solves all of them, and it is easy to set up the wrong way round. This post explains the gap method, works it on people, repeated letters and seat selection, and shows why NCERT's "never together" answer is a different question.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists permutations and combinations with "Simple applications", and the JEE (Advanced) 2026 brochure includes permutations and combinations and their use in probability. NCERT's Class 11 chapter Permutations and Combinations (2026-27 reprint) uses the gap method in its Example 24, seating 5 girls and 3 boys so that no two boys are together. Check the current documents for your exam.
How does the gap method work?
Arrange the free items first, the ones with no restriction.
Count the gaps. k items in a row leave k + 1 gaps: one before the first, one between each pair, one after the last.
Choose gaps for the restricted items, at most one per gap, so no two can touch.
Arrange the restricted items in the chosen gaps if they are different from one another.
Multiply.
NCERT's Example 24 draws this as × G × G × G × G × G ×: five girls leave six marked places for the three boys.
The order matters. If you place the restricted items first and then try to keep the others apart, you end up counting cases by hand.
Worked example 1: six boys and four girls
In how many ways can 6 boys and 4 girls stand in a row so that no two girls are together?
Arrange the boys: 6! = 720.
Six boys leave 7 gaps.
Choose 4 of the gaps and arrange the girls in them: ⁷P₄ = 7 × 6 × 5 × 4 = 840.
Total: 720 × 840 = 604,800.
Quick sanity check with smaller numbers: 3 boys and 2 girls give 3! × ⁴P₂ = 6 × 12 = 72, which you can confirm by listing patterns such as G B G B B.
If there are too few gaps, the answer is zero. Five girls cannot be kept apart by three boys, because three boys leave only four gaps.
Is "no two together" the same as "not all together"?
No, and this is where many answers go wrong. With the same 6 boys and 4 girls:
All four girls together: treat them as one block. 7 objects in 7! = 5,040 ways, times 4! = 24 orders inside the block: 120,960.
Not all together: 10! − 120,960 = 3,628,800 − 120,960 = 3,507,840.
No two together: 604,800, from the gap method.
"Not all together" includes rows where two girls stand side by side, as long as the four are not in one block. "No two together" forbids every pair. Read the condition twice.
Worked example 2: repeated letters
How many arrangements of INDEPENDENCE have no two vowels next to each other?
INDEPENDENCE has 12 letters. The vowels are E, E, E, E, I (five). The consonants are N, N, N, D, D, P, C (seven).
Arrange the consonants: 7!/(3! × 2!) = 5040/12 = 420.
Seven consonants leave 8 gaps. Choose 5 for the vowels: C(8, 5) = 56.
Arrange the vowels in those gaps: 5!/4! = 5, because the four E's are identical.
Total: 420 × 56 × 5 = 117,600.
Compare NCERT's answer for "never together" in its sense: 1,663,200 − 16,800 = 1,646,400. A computer count of all 1,663,200 arrangements confirms both numbers, 117,600 with no two vowels adjacent and 1,646,400 without the five vowels in one block. The words are close; the counts are not.
Worked example 3: choosing seats
In how many ways can 3 seats be chosen from a row of 10 so that no two chosen seats are adjacent?
Think of the 7 seats you do not choose as the free items. They leave 8 gaps, and each chosen seat goes into a different gap. The seats are just positions, so choose the gaps without arranging: C(8, 3) = 56. A full listing of all C(10, 3) = 120 choices finds exactly 56 with no neighbours.
If three different people then sit in those seats, multiply by 3! = 6 to get 336.
Which mistakes cost marks?
Placing the restricted items first.
Counting k − 1 gaps and forgetting the two ends.
Using C instead of P when the restricted items are different, or P when they are identical.
Treating "not all together" as "no two together".
Using the gap method on a circle without changing the count: n people round a table leave n gaps, not n + 1. Our post on circular permutations covers that case.
A practice routine
Solve NCERT Example 24 yourself, then change it to 5 boys and 3 girls with no two girls together.
Redo example 2 for MISSISSIPPI with no two S's adjacent. You should get 7!/(4! 2!) × C(8, 4) = 105 × 70 = 7,350. Our post on arranging letters with repeats has more on that word.
Mix gap problems with block problems in one set. Rohrer and Taylor (2007) found that students who practised mixed problem types did far better on a test a week later than those who practised one type at a time (ERIC abstract).
When you cannot tell whether a problem wants a block or the gaps, the Fermi Problem Library has a step-by-step solver that explains the reasoning behind each choice.
Today, count the arrangements of the letters of ALGEBRA with no two vowels together, and check it by a second method.
