JEE Corner — Permutations and Combinations for JEE Main and Advanced

What JEE Adds

The textbook chapter gives you the four formulas — the counting principle, $^nP_r$, repeated-letter division and $^nC_r$. JEE Main tests those plus a small toolkit of standard moves: circular arrangements, the bundle-gap-complement trio for adjacency restrictions, dictionary rank, any-size selections ($2^n$), and case-split at-least/at-most counts. Every one of them is a two-line extension of what you already know — and every one appears in past papers year after year.

Key Point: JEE counting problems are rarely hard; they are precise. The mark is lost in reading — "at least" vs "exactly", "not all together" vs "no two together", "numbers" vs "strings with leading zeros". Slow down on the sentence, not the arithmetic.

Circular Permutations

Around a round table, rotating everyone one seat produces the same seating — only relative order matters. Fix one person as an anchor and arrange the remaining $n - 1$ linearly:

Key Point: $n$ distinct objects in a circle: $(n-1)!$ arrangements. If clockwise and anticlockwise are indistinguishable (necklaces, garlands): $\frac{(n-1)!}{2}$.

Standard variants:

• 5 people at a round table: $4! = 24$.
• Garland of 6 different flowers: $\frac{5!}{2} = 60$.
• Couple must sit together at a round table of 6: bundle the couple — 5 units in a circle: $4!$, times $2!$ inside: $48$.
• 4 men and 4 women alternating at a round table: seat the men circularly ($3! = 6$), then the women in the 4 gaps between them ($4! = 24$): $144$.

[JEE Tip] Circular formulas apply to unnumbered seats. If the chairs are numbered (or one seat is special), rotations become distinguishable and the count returns to plain $n!$.

Bundle, Gap, Complement — and Rank

The three adjacency moves, in one line each:

• Together (bundle): glue the group into one unit; multiply by internal arrangements. DAUGHTER vowels together: $6! \times 3! = 4320$.
• Not all together (complement): total minus together: $8! - 4320 = 36000$.
• No two adjacent (gaps): arrange the free objects, then place the restricted ones in the gaps. 5 girls, 3 boys, no two boys adjacent: $5! \times {^6P_3} = 14400$. Gaps are the only correct tool here — the complement of "all together" is weaker.

Dictionary rank: to find the position of a word among all arrangements of its letters listed alphabetically, count the complete blocks that precede it — words starting with an earlier letter, then within the block, words with an earlier second letter, and so on.

Any-size selections: each of $n$ distinct objects is either in or out — $2^n$ subsets, $2^n - 1$ if at least one object must be chosen. For "at least one" questions on identical objects or with conditions, complement counting (all minus none) is almost always the shortest route.

[JEE Tip] $^nC_0 + {^nC_1} + \cdots + {^nC_n} = 2^n$ — the selection-of-any-size identity — reappears as a binomial identity next chapter. Quote it, don't re-derive it.

Solved Examples

Example 1: Round table basics

In how many ways can 5 friends sit around a circular table?

Solution:

Step 1 — Break the rotation symmetry. Around an unmarked round table, rotating everyone one seat gives the same seating. Fix one friend as an anchor.

Step 2 — Arrange the rest linearly. The remaining 4 friends fill the seats relative to the anchor: $4! = 24$.

Takeaway: Circle of $n$ distinct people = $(n-1)!$ — anchor one, arrange the rest.

Example 2: Garland

How many different garlands can be made from 6 different flowers?

Solution:

Step 1 — Circular count. As a circle: $(6-1)! = 5! = 120$.

Step 2 — Kill the flip. A garland looks the same turned over, so clockwise and anticlockwise versions coincide: divide by 2.

Step 3 — Compute. $\frac{120}{2} = 60$ garlands.

Takeaway: Necklaces and garlands are circles that can flip: $\frac{(n-1)!}{2}$.

Example 3: Couple together, round table

In how many ways can 6 people sit at a round table if two particular people must sit together?

Solution:

Step 1 — Bundle the pair. Glue the two into one unit — now 5 units sit in a circle: $(5-1)! = 4! = 24$.

Step 2 — Open the bundle. The pair swaps internally: $\times 2! = 2$.

Step 3 — Multiply. $24 \times 2 = 48$ seatings.

Takeaway: The bundle method works in circles too — just apply $(n-1)!$ to the reduced unit count.

Example 4: Alternating men and women

In how many ways can 4 men and 4 women sit alternately around a round table?

Solution:

Step 1 — Seat one group circularly. The 4 men around the table: $(4-1)! = 3! = 6$.

Step 2 — Fill the gaps. The men create exactly 4 gaps between them; the 4 women fill them linearly (the anchor is already fixed): $4! = 24$.

Step 3 — Multiply. $6 \times 24 = 144$.

Takeaway: No extra factor of 2 — swapping "men's seats" and "women's seats" is just a rotation of the same seating.

Example 5: Circle vs numbered chairs

Eight delegates sit (i) around an unmarked round table (ii) on 8 numbered chairs in a circle. Count the seatings.

Solution:

Step 1 — (i) Unmarked table. Rotations are identical, so anchor one delegate: $(8-1)! = 7! = 5040$.

Step 2 — (ii) Numbered chairs. Chair numbers make every rotation a different assignment — the symmetry is gone: $8! = 40320$.

Step 3 — Check the ratio. $40320 = 8 \times 5040$ ✓ — each circular seating splits into 8 distinguishable rotations.

Takeaway: $(n-1)!$ belongs to unnumbered seats only; numbering any seat restores $n!$.

Example 6: No two alike adjacent — gaps with repeats

In how many arrangements of the word MISSISSIPPI do no two I's appear together?

Solution:

Step 1 — Arrange the non-I letters. M, four S's and two P's: $\frac{7!}{4!\,2!} = \frac{5040}{48} = 105$.

Step 2 — Count the gaps. Seven letters in a row create 8 gaps (including both ends).

Step 3 — Place the identical I's. Choose 4 of the 8 gaps — combinations, because the I's are identical: $^8C_4 = 70$.

Step 4 — Multiply. $105 \times 70 = 7350$. (For contrast, the weaker condition "the four I's are not all together" gives 33810 — a different question.)

Takeaway: Gap method with identical movers uses $C$, not $P$ — ordering identical letters would overcount.

Example 7: Rank of a word

Find the rank of the word RACE among all arrangements of its letters in dictionary order.

Solution:

Step 1 — Alphabetise the letters. A, C, E, R.

Step 2 — Count whole blocks before the R-block. Words starting with A, C or E: $3 \times 3! = 18$.

Step 3 — Locate RACE inside the R-block. After R, the smallest completion is A-C-E — which is exactly RACE, the first R-word.

Step 4 — Add. Rank $= 18 + 1 = 19$.

Takeaway: Rank = (words strictly before it) + 1 — never forget the final +1.

Example 8: Rank with repeated letters

Find the rank of AGAIN in the dictionary listing of its own arrangements.

Solution:

Step 1 — Alphabetise. Letters A, A, G, I, N — first letters available: A, G, I, N.

Step 2 — Words before the A-G block. AGAIN starts A-G. Earlier second letters: only A (from A-A…): arrange the remaining G, I, N freely: $3! = 6$ words.

Step 3 — Inside the A-G block. Third letter of AGAIN is A — the smallest available; A-G-A words arrange I, N: AGAIN then AGANI. AGAIN is the very first.

Step 4 — Add. Rank $= 6 + 1 = 7$.

Takeaway: With repeated letters, count remaining stock carefully at each position — the two A's are identical.

Example 9: Any-size selections

A shop stocks 8 different books. In how many ways can a customer buy at least one book?

Solution:

Step 1 — In-or-out per book. Each of the 8 distinct books is independently bought or not: $2^8 = 256$ subsets.

Step 2 — Exclude the empty purchase. "At least one" removes exactly one subset: $256 - 1 = 255$.

Takeaway: Distinct objects, any size: $2^n$; at least one: $2^n - 1$.

Example 10: At least one of each type — complement

From 5 different maths and 4 different physics books, how many selections contain at least one book of each subject?

Solution:

Step 1 — Nonempty choice per subject. Maths: $2^5 - 1 = 31$; physics: $2^4 - 1 = 15$.

Step 2 — Multiply the independent choices. $31 \times 15 = 465$.

Step 3 — Why not $2^9 - 1$? That count ($511$) allows all-maths or all-physics selections, which break the "one of each" condition.

Takeaway: Apply the nonempty correction per pool, then multiply.

Example 11: At least one ace

How many 5-card hands from 52 cards contain at least one ace?

Solution:

Step 1 — Complement. "At least one ace" = all hands − hands with no ace.

Step 2 — Count both. All: $^{52}C_5 = 2598960$; ace-free (choose from 48 non-aces): $^{48}C_5 = 1712304$.

Step 3 — Subtract. $2598960 - 1712304 = 886656$.

Takeaway: Never compute "at least one" by cases when the complement is a single subtraction. (And $^4C_1 \times {^{51}C_4}$ is wrong — it counts multi-ace hands more than once.)

Example 12: Exactly vs at least — the contrast

From 6 men and 5 women, count committees of 4 with (i) exactly 2 women (ii) at least 2 women.

Solution:

Step 1 — (i) Exactly 2 women. $^5C_2 \times {^6C_2} = 10 \times 15 = 150$.

Step 2 — (ii) Add the higher cases. Women $= 3$: $^5C_3 \times {^6C_1} = 10 \times 6 = 60$; women $= 4$: $^5C_4 \times {^6C_0} = 5 \times 1 = 5$.

Step 3 — Total. $150 + 60 + 5 = 215$.

Takeaway: "At least $k$" = sum of the "exactly" cases from $k$ up to the stock limit.

Example 13: Total selections from mixed identical objects

A basket has 4 identical apples and 3 identical oranges. How many non-empty selections of fruit are possible?

Solution:

Step 1 — Identical objects are chosen by count only. Apples: take 0, 1, 2, 3 or 4 — that is $4 + 1 = 5$ ways. Oranges: $3 + 1 = 4$ ways.

Step 2 — Multiply and remove the empty case. $5 \times 4 - 1 = 19$.

Takeaway: Identical stock contributes $(p+1)$ options per type — the $2^n$ formula is for distinct objects only.

Example 14: Sum of all numbers formed

Find the sum of all 4-digit numbers formed using digits 1, 2, 3, 4 exactly once each.

Solution:

Step 1 — Count appearances per place. There are $4! = 24$ numbers; by symmetry each digit occupies each place $\frac{24}{4} = 6$ times.

Step 2 — Sum one place. Column total: $6 \times (1 + 2 + 3 + 4) = 60$ in each of units, tens, hundreds, thousands.

Step 3 — Assemble by place value. $60 \times (1000 + 100 + 10 + 1) = 60 \times 1111 = 66660$.

Takeaway: The digit-sum trick: (appearances per place) × (digit sum) × ($111\ldots1$).

Example 15: Choosing with a forbidden pair

From 10 people including A and B, how many committees of 4 contain neither A nor B together — i.e. A and B are never BOTH on the committee?

Solution:

Step 1 — Complement. Count committees with BOTH A and B: fix them, choose 2 more from the other 8: $^8C_2 = 28$.

Step 2 — Subtract from the total. $^{10}C_4 - 28 = 210 - 28 = 182$.

Step 3 — Sanity check by cases. Neither: $^8C_4 = 70$; exactly one of A, B: $2 \times {^8C_3} = 2 \times 56 = 112$; $70 + 112 = 182$ ✓.

Takeaway: A forbidden pairing is one clean complement — and a case count confirms it in ten seconds.