Permutations and Combinations

Arrangements and selections

A permutation counts arrangements, where order matters: the medals gold, silver, bronze are different outcomes for the same three runners. A combination counts selections, where order doesn't matter: a committee of the same three people is one committee whichever order you name them in.

Ask one question before using any formula: if I swap two of the chosen items, do I get a different outcome? If yes, count permutations. If no, count combinations.

The multiplication principle

If one choice can be made in $m$ ways and, after it, a second choice in $n$ ways, the two together can be made in $m \times n$ ways. Every formula on this page comes from this one rule.

Arranging 5 books on a shelf. 5 choices for the first position, 4 for the second, and so on: $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$ arrangements.

Codes. A 3-letter code from the 26 letters with no letter repeated: $26 \times 25 \times 24 = 15{,}600$. If letters may repeat, every position has 26 choices: $26^3 = 17{,}576$.

The formulas

Choosing $r$ items from $n$ distinct items:

The last identity says it all: to arrange $r$ items, first choose which $r$ (${}^nC_r$ ways), then arrange them ($r!$ ways). For example ${}^{10}P_3 = 720$ and ${}^{10}C_3 = 120$, and $120 \times 3! = 720$.

Worked example: a committee

From 5 men and 4 women, how many committees of 3 have exactly 2 men?

Solution: Choose the 2 men (${}^5C_2 = 10$ ways), then the 1 woman (${}^4C_1 = 4$ ways). By the multiplication principle, $10 \times 4 = 40$ committees.

At least one woman. Counting "at least one" case by case is slow. Count everything and subtract the committees with no women: ${}^9C_3 - {}^5C_3 = 84 - 10 = 74$.

Worked example: identical letters

How many different arrangements are there of the letters of BANANA?

Solution: 6 letters give $6!$ orderings, but swapping the three A's among themselves, or the two N's, produces the same word. Divide out those repeats:

Worked example: seats around a table

In how many ways can 6 people sit around a round table, if only their positions relative to each other matter?

Solution: Rotating everyone one seat gives the same seating, so fix one person's seat and arrange the other 5: $5! = 120$. In general, $n$ people in a circle can be arranged in $(n-1)!$ ways. If the arrangement can also be flipped over, like beads on a necklace, divide by 2 more.

Worked example: choose, then arrange

How many 4-letter strings with all letters different can be made using exactly 2 of the 5 vowels and 2 of the 21 consonants?

Solution: Choose the vowels (${}^5C_2 = 10$), choose the consonants (${}^{21}C_2 = 210$), then arrange the 4 chosen letters ($4! = 24$):

Mixing "choose" and "arrange" in one step is where most errors come from. Separate them: select first with combinations, then arrange the selection.

Worked example: the 19th word in dictionary order

All arrangements of the letters of FROG are listed in dictionary order. Which is the 19th?

Solution: Sort the letters: F, G, O, R. Each starting letter leads a block of $3! = 6$ words, so words 1 to 6 start with F (FROG itself is the 6th), 7 to 12 with G, 13 to 18 with O, and 19 to 24 with R. The 19th word is the first word starting with R, with the rest in alphabetical order: RFGO.

Common mistakes

• Using permutations for a committee. A committee, a team or a hand of cards is a selection. If swapping two members gives the same outcome, it's ${}^nC_r$.
• Forgetting to divide for identical items. BANANA has 60 arrangements, not $6! = 720$.
• Counting rotations as different around a table. $n$ people in a circle give $(n-1)!$ arrangements, not $n!$.
• Adding where you should multiply. Independent choices made one after the other multiply. Add only when the cases are separate alternatives (this committee or that one).
• Counting "at least one" case by case and double-counting. Use the complement: total minus the cases with none.

Practice questions

1. Seven runners finish a race. In how many ways can gold, silver and bronze be awarded?
2. In how many ways can a team of 4 be chosen from 12 players?
3. How many different arrangements are there of the letters of LEVEL?
4. From 6 boys and 4 girls, how many groups of 4 contain exactly 2 girls?
5. In how many ways can 5 people sit around a round table?
6. From a class of 6 boys and 4 girls, how many groups of 4 contain at least one girl?

Answers

1. Order matters: ${}^7P_3 = 7 \times 6 \times 5 = 210$.
2. Order doesn't matter: ${}^{12}C_4 = 495$.
3. 5 letters with two L's and two E's: $\dfrac{5!}{2!\;2!} = 30$.
4. ${}^6C_2 \times {}^4C_2 = 15 \times 6 = 90$.
5. $(5-1)! = 24$.
6. All groups minus the all-boy groups: ${}^{10}C_4 - {}^6C_4 = 210 - 15 = 195$.

Where this comes up in exams

JEE Main 2026 and JEE Advanced 2026 both list permutations and combinations, and it is a Class XI chapter in the 2025-26 CBSE syllabus. The General Aptitude section that every GATE 2027 paper includes lists permutations and combinations under numerical computation and estimation.

Checked against JEE (Main) 2026 syllabus (NTA), JEE (Advanced) 2026 syllabus, CBSE Mathematics curriculum, Classes XI-XII (2025-26) and GATE 2027 information brochure (General Aptitude syllabus). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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