Why this topic matters · 8 min read
Permutation and Combination appears in SBI PO Prelims and Mains under Quantitative Aptitude, typically 2-3 questions per exam. Questions range from straightforward arrangement problems to tricky probability-linked combo problems in Mains. SBI PO tends to test word arrangements, selection from groups, circular arrangements, and committee formation. Mains may club this with probability. Scoring here is high if you lock down the nPr vs nCr decision quickly.
Core Idea: Permutation vs Combination
Permutation is arrangement — ORDER MATTERS. Combination is selection — ORDER DOES NOT MATTER. Think of it this way: if you are picking a cricket team (11 players), order does not matter — that is Combination. If you are deciding who bats at which position, order matters — that is Permutation. The single most important decision in any P and C problem is: does order matter or not?
- Permutation = Arrangement = Order matters
- Combination = Selection = Order does not matter
- nPr is always greater than or equal to nCr for same n and r
- nCr = nC(n-r) — very useful shortcut for large values
- nC0 = nCn = 1, nC1 = n
- nPr = nCr x r! — Permutation is just Combination multiplied by the arrangements of chosen items
Key formulas
Permutation formula
nPr = n! / (n - r)!
When: Use when arranging r items out of n distinct items where order matters
Combination formula
nCr = n! / (r! x (n - r)!)
When: Use when selecting r items out of n distinct items where order does not matter
Relation between P and C
nPr = nCr x r!
When: Use to convert between selection and arrangement quickly
Worked examples
How many ways can 3 people be selected from a group of 7? Order does not matter (selection), so use 7C3 = 7! / (3! x 4!) = (7x6x5) / (3x2x1) = 35 ways.
How many 3-letter words (with or without meaning) can be formed from 7 distinct letters? Order matters (arrangement), so use 7P3 = 7! / 4! = 7x6x5 = 210 ways.
Factorial Basics
Factorial is the building block. n! means multiply all integers from 1 to n. Remember: 0! = 1 (this is a definition, not a derivation — just memorize it). Large factorials cancel nicely — never expand fully, always cancel from the top.
- 0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040
- Cancel factorials top and bottom before multiplying — saves time
- n! = n x (n-1)!
- Factorials grow very fast — 10! = 3628800
Key formulas
Factorial definition
n! = n x (n-1) x (n-2) x ... x 2 x 1
When: Base of all P and C calculations
Arrangement with Restrictions
SBI PO loves restriction-based problems. The key strategies are: fix the restricted items first, then arrange the rest. If certain items MUST be together, treat them as one unit (bundle trick). If certain items MUST NOT be together, use total arrangements minus the cases where they ARE together.
- Items must be together: Tie them into one bundle, arrange (n-k+1) items, then arrange k items within the bundle
- Items must NOT be together: Total arrangements minus arrangements where they are together
- Vowels and consonants problems: Fix consonants first, then place vowels in gaps
- For MISSISSIPPI type words: n! divided by (repeat1! x repeat2! x ...) for repeated letters
- Always subtract restricted cases from total when it is easier
Key formulas
Arrangements with identical items
n! / (p! x q! x r!)
When: When a word or sequence has repeated elements: p, q, r are counts of each repeated item
Bundle trick
(n - k + 1)! x k!
When: When k specific items must always stay together out of n total items
Worked examples
Word BANANA: 6 letters, A repeats 3 times, N repeats 2 times, B once. Arrangements = 6! / (3! x 2! x 1!) = 720 / (6x2) = 60.
In how many ways can 5 people sit in a row if 2 specific people must sit together? Bundle the 2 as one unit: 4 units arrange in 4! = 24 ways, the 2 within the bundle arrange in 2! = 2 ways. Total = 24 x 2 = 48.
Circular Arrangement
In a circular arrangement, there is no fixed starting point. So we fix one person and arrange the rest. This removes the rotational duplicates. For a necklace or bracelet (can be flipped), divide further by 2 because clockwise and anticlockwise look the same.
- Circular arrangements of n people = (n-1)!
- Necklace or bracelet = (n-1)! / 2
- Fix one person at top, arrange remaining n-1 in (n-1)! ways
- SBI PO often asks circular seating with restrictions — apply bundle trick after fixing reference point
Key formulas
Circular permutation
(n - 1)!
When: Arranging n distinct people around a circular table
Necklace arrangement
(n - 1)! / 2
When: Circular arrangement where clockwise and anticlockwise are same (jewellery, garlands)
Worked example
8 people sit around a round table. Ways = (8-1)! = 7! = 5040.
Selection Problems: Committee and Teams
These are pure Combination problems. SBI PO frequently asks: select a committee of X from Y men and Z women with some condition (at least 2 women, exactly 1 man, etc.). The approach: break into cases satisfying the condition, calculate each case using nCr, then add all cases.
- At least one: Total selections minus selections with none of that type
- Exactly k: Direct nCr selection for that specific count
- At least 2 women in a 4-member committee: Add cases of exactly 2 women + exactly 3 women + exactly 4 women
- When two people must always be included: Select remaining from rest
- When two people cannot both be included: Total minus cases where both are selected
Key formulas
At least one shortcut
Total selections - selections with zero of that category
When: Faster than adding all individual cases when at least one condition applies
Worked example
A committee of 4 is formed from 6 men and 4 women. How many ways if at least 1 woman must be included? Total ways = 10C4 = 210. Ways with no woman = 6C4 = 15. Answer = 210 - 15 = 195.
⚠ Common mistakes to avoid
- Confusing when to use nPr vs nCr: Always ask yourself first — does order matter? Seating arrangements = P, team selection = C.
- Forgetting to divide by repeat factorials in word arrangement problems like MISSISSIPPI, BANANA, LEVEL.
- In circular arrangements, using n! instead of (n-1)! — the most common error in seating-around-a-table problems.
- In at-least problems, students try to add all cases manually instead of using the shortcut: Total minus (none of that type). This wastes time and increases error.
- Misreading the question — SELECTED vs ARRANGED. Committee of 5 selected is C; then if they are given specific roles it becomes P on top of C.
🧠 Memory aids
- PARty needs ORDER — Permutation = Arrangement = Order matters. CAS (Choose A Squad) — Combination = Selection = no order.
- Circular table? Fix one, free the rest: (n-1)! — think of one person nailed to a chair.
- SAME letters? SAME factorial in denominator. BANANA has 3 As so 3! below, 2 Ns so 2! below.
- AT LEAST shortcut = TOTAL minus NONE. This is the Swiss Army knife of P and C — use it whenever you see the words at least or at most with one boundary.
🎯 SBI PO exam tips
- SBI PO Prelims usually has 1-2 direct formula-based questions (word arrangement or committee selection) — solvable in under 60 seconds if formulas are memorized.
- SBI PO Mains often combines P and C with Probability — set up the combination count first, then form the probability fraction. Practice this dual-step approach.
- Restriction-based questions (must sit together, cannot sit together) appear frequently in Mains DI sets disguised as seating arrangement quant problems — stay alert.
- For word arrangement questions in SBI PO, check for repeated letters immediately before applying any formula. Skipping this costs a direct wrong answer.
- Time target: Prelims P and C questions should be solved in 45-75 seconds. If a question takes more than 90 seconds, mark and move — do not let one question derail the section.
Q1 · medium · AI-verified
In how many ways can 8 people be arranged in a row such that 3 particular people are always together?
- 2880
- 3600
- 4320
- 5040
Q2 · medium · AI-verified
In how many ways can 8 people be seated around a circular table such that 2 particular people always sit together?
- 5040
- 2880
- 1440
- 720
Q3 · medium · AI-verified
In how many ways can 10 identical balls be distributed among 4 distinct boxes such that no box remains empty?
- 84
- 120
- 126
- 165
Q4 · medium · AI-verified
From a group of 8 boys and 6 girls, a team of 11 players is to be selected. In how many ways can this be done if the team must include exactly 6 boys and 5 girls?
- 126
- 168
- 252
- 210
Q5 · medium · AI-verified
How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6 without repetition such that the number is divisible by 4?
- 60
- 72
- 84
- 96