Permutation & Combination • Topic 4 of 4

Restricted Arrangements

Restrictions are where CAT separates fast solvers from the rest, and two methods cover most cases. The grouping (or "string together") method handles "must be together": glue the linked items into a single block, arrange the resulting units, then arrange inside the block. For "always together", n items with a fixed pair give (n−1)! × 2!. The gap method handles "must be apart" or "no two together": first arrange the unrestricted items, then drop the restricted items into the gaps between them (including the ends). If k items must all be separated among m others, arrange the m in m! ways, creating m+1 gaps, and place the k items in those gaps in (m+1)Pk ways. A third tool is the complement: count "at least one pair together" as total minus "none together". Repetition still applies — divide by the factorials of repeated letters when the items are not all distinct.

✅ Solved examples

1. How many arrangements of the letters of BANANA?
6 letters with A repeated 3 times and N repeated 2 times: 6!/(3! 2!) = 720/12 = 60.
2. In how many ways can the letters of LEADING be arranged so that the vowels (E, A, I) are always together?
Glue EAI into one block ⇒ units are L, D, N, G and the block = 5 units ⇒ 5! = 120; the 3 vowels inside arrange in 3! = 6. Total = 120 × 6 = 720.
3. In how many ways can 4 boys and 3 girls sit in a row so that no two girls sit together?
Seat the 4 boys first: 4! = 24, creating 5 gaps. Place 3 girls in the gaps: 5P3 = 60. Total = 24 × 60 = 1,440.
4. How many arrangements of the word ARRANGE have both R’s never together?
Total arrangements of ARRANGE = 7!/(2! 2!) = 1260 (A and R each repeat twice). With both R’s together, glue them: treat RR as one ⇒ 6!/(2!) = 360. Never together = 1260 − 360 = 900.

✏️ Practice — try these, take hints as needed

1. How many distinct arrangements of the letters of LEVEL?
5 letters: L twice, E twice.
5!/(2! 2!).
120/4.
30
2. In how many ways can the letters of NUMBER be arranged so the vowels (U, E) are always together?
Glue UE into one block.
5 units ⇒ 5!.
Times 2! for the two vowels.
240
3. In how many ways can 3 boys and 3 girls sit in a row so that boys and girls alternate?
Two patterns: BGBGBG or GBGBGB.
Each = 3! × 3!.
Add the two patterns.
72
4. How many arrangements of MISSISSIPPI?
11 letters: I×4, S×4, P×2, M×1.
11!/(4! 4! 2!).
39,916,800 / (24 × 24 × 2).
34,650
5. In how many ways can 5 men and 2 women sit in a row so the 2 women are never together?
Seat 5 men: 5!, giving 6 gaps.
Place 2 women in gaps: 6P2.
120 × 30.
3,600

📝 Topic test — 8 questions

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