Permutations and Combinations

2.1 Core concepts

• Counting techniques replace brute-force listing: enumerating all arrangements of 9 remaining digits taken 3 at a time would be tedious, so counting principles avoid enumeration (NCERT §6.1, p. 100). Explicit listing is impractical even for moderate inputs — use multiplication and factorial reasoning instead.
• Fundamental Principle of Counting (Multiplication Principle): If an event can occur in m different ways and following it another event can occur in n different ways, then the two events in the given order can occur in m × n ways; this generalises to any finite number of events (NCERT §6.2, p. 102). Every permutation and combination formula follows from this single principle.
• Illustrative applications: 3 pants × 2 shirts = 6 pant–shirt pairs; 2 bags × 3 tiffin boxes × 2 water bottles = 12 ways to carry items (NCERT §6.2, pp. 100–101). The same logic powers digit-formation, code-generation, and seating problems.
• Permutation (Definition 1): an arrangement in a definite order of a number of objects taken some or all at a time (NCERT §6.3, p. 105). The phrase "definite order" is what distinguishes permutations from combinations.
• Theorem 1 — Permutations of n distinct objects taken r at a time (no repetition): nPr = n(n-1)(n-2)...(n-r+1), proved by filling r vacant places successively (NCERT §6.3.1, p. 105). The first place can be filled n ways, the second (n − 1) ways, …, the r-th (n − r + 1) ways; multiply by the fundamental principle.
• Factorial notation: n! = 1 × 2 × 3 × ... × (n-1) × n; also n! = n · (n-1)!. By definition 0! = 1 (NCERT §6.3.2, p. 105). The recursion n! = n(n−1)! lets one extend the definition backwards and underwrites the convention 0! = 1.
• Closed-form for nPr: nPr = n!/(n-r)!, valid for 0 ≤ r ≤ n. In particular nPn = n!/0! = n! and nP0 = 1 (NCERT §6.3.3, pp. 107–108). The 0! = 1 convention makes these edge cases consistent.
• Theorem 2 — Permutations with repetition allowed: number of permutations of n different objects taken r at a time with repetition is n^r (NCERT §6.3.3, p. 108). Example: 3-letter "words" from NUMBER with repetition = 6^3 = 216.
• Theorem 3 (special case): number of permutations of n objects where p are of the same kind and the rest are all different = n!/p! (NCERT §6.3.4, p. 110). The division by p! removes overcounting from the indistinguishability of the p identical objects.
• Theorem 4 (general case): number of permutations of n objects where p1 are of one kind, p2 of a second kind, ..., pk of a kth kind, and the rest are all different = n!/(p1! p2! ... pk!) (NCERT §6.3.4, p. 110). Worked: ALLAHABAD has 9 letters with 4 A's and 2 L's, giving 9!/(4! 2!) = 7560 (Example 9, p. 110).
• Combination: a selection of objects where order does not matter; for example, from players X, Y, Z only 3 two-player teams exist: XY, YZ, ZX (NCERT §6.4, pp. 114–115). The same 3 teams correspond to 6 ordered pairs (XY ≠ YX as permutations) — illustrating r! collapse.
• Theorem 5 — Relationship between permutations and combinations: nPr = nCr · r! for 0 < r ≤ n, because each combination of r objects can be rearranged in r! ways (NCERT §6.4, p. 116). This is the conceptual bridge — combinations first, then internally permute.
• Closed-form for nCr: nCr = n!/(r!(n-r)!), 0 ≤ r ≤ n. By convention nC0 = 1 and nCn = 1 (NCERT §6.4, p. 116). The symmetry between r and n − r is built into the formula.
• Identity nCr = nC(n-r): selecting r objects out of n is the same as rejecting (n-r) objects; also nCa = nCb ⇒ a = b or a + b = n (NCERT §6.4, p. 117). This dual rule is heavily tested.
• Theorem 6 (Pascal's identity): nCr + nC(r-1) = (n+1)Cr (NCERT §6.4, p. 117, proof on p. 117). Equivalently, the binomial-coefficient Pascal triangle: each entry is the sum of the two above it.
• Applications worked in chapter: choosing a Chairman and Vice-Chairman from 12 persons = 12P2 = 132 (p. 108); 4-card selections from a 52-card deck = 52C4 = 270725 with breakdowns by suit / colour / face cards (Example 19, pp. 117–119); committee of 3 from 2 men and 3 women = 5C3 = 10, of which 1-man-2-women = 2C1 × 3C2 = 6 (Example 18, p. 117); team of 5 from 4 girls + 7 boys with at-least-one-of-each = 441, with at-least-3-girls = 91 (Example 21, p. 120).
• Recognising the structure of the problem (arrangement vs selection, repetition allowed or not, identical objects present, restrictions on positions) matters more than memorising formulas.
• Special tally for "no two adjacent" constraints: arrange the non-restricted items first; this creates n + 1 "gaps" (including the ends) into which the restricted items can be placed without adjacency. The count is then (gap choices)C(restricted count) × arrangements of restricted within their chosen gaps.
• A useful sanity check on every count: the answer must be a positive integer. Negative or non-integer results signal a formula error or a forgotten case.
• The "stars and bars" interpretation (not in NCERT but implicit in some Misc. Ex. questions): number of non-negative integer solutions of x₁ + x₂ + … + xₖ = n equals (n + k − 1)C(k − 1). This generalises the combination idea to "selection with repetition".
• Card-deck counting follows directly from the combination formula nCr: 52 cards split into 4 suits of 13; faces (J, Q, K) total 12 across the deck; choosing r cards from a particular subset uses nCr on that subset's size.

2.7 Worked permutation/combination contrast

NCERT carefully contrasts permutations and combinations in §6.4 (p. 115). Consider the 4 letters A, B, C, D taken 3 at a time. Permutations: 4P3 = 24 (e.g., ABC, ACB, BAC, BCA, CAB, CBA are six distinct permutations of the same three letters). Combinations: 4C3 = 4 ({A,B,C}, {A,B,D}, {A,C,D}, {B,C,D}). The ratio 24/4 = 6 = 3! confirms the bridge identity nPr = nCr · r!. CUET often tests this distinction by inserting the words "arrange" vs "choose" in stems that are otherwise identical.

A second illustration uses number-formation problems. From digits {1, 2, 3, 4, 5}: (i) number of 3-digit numbers without repetition = 5P3 = 60 (ordered); (ii) number of 3-element subsets of {1,…,5} = 5C3 = 10 (unordered). The ratio again confirms 60 = 10 · 3!. Whenever a problem refers to "numbers" or "codes" the order matters; when it refers to a "group" or "selection", order does not. Read the stem carefully and underline the key verb to decide which formula applies.

A third illustration: from 6 men and 4 women, choose a committee of 5 with at least 2 women. Approach: split by exactly 2, 3, 4 women.

• Exactly 2 women: 4C2 · 6C3 = 6 · 20 = 120.
• Exactly 3 women: 4C3 · 6C2 = 4 · 15 = 60.
• Exactly 4 women: 4C4 · 6C1 = 1 · 6 = 6. Total = 186. This case-split structure recurs in nearly every "at least" CUET word problem.

2.8 Quick computational shortcuts

• nC1 = n; nC2 = n(n−1)/2; nC3 = n(n−1)(n−2)/6. Memorising these three saves time on many MCQs.
• When n is small (≤ 10) and r is small, expand nCr as the product (n)(n−1)…(n−r+1) divided by r!; never compute the full factorials.
• For nCr with r > n/2, use symmetry: nCr = nC(n−r). E.g., 12C9 = 12C3 = 220.
• Always check whether the answer-choice list has only one positive integer; combinatorics options are sometimes given with one negative or fractional distractor that is impossible.
• For "letters together" problems, the block count after gluing is (n − k + 1)! where k is the size of the block; then multiply by k! for the internal arrangements.
• For "letters never together" use total − together (complement); for "no two together" use the gaps method.
• For "starts with X and ends with Y" fix the two ends first, then permute the rest in the middle.
• For divisibility constraints on the formed number, fix the last digit (or last two digits) per the divisibility rule first, then count permutations of remaining digits in remaining places. For divisibility by 5, the last digit must be 0 or 5; for divisibility by 2, even; for 4, last two divisible by 4; for 3 or 9, digit sum divisible by 3 or 9 respectively. These number-theoretic constraints cascade neatly into multiplication-principle counts.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig. 6.1 (p. 101): Tree-style diagram pairing 3 pants P1, P2, P3 with 2 shirts S1, S2 to give 6 outfit combinations — visual proof of the multiplication principle.
• Fig. 6.2 (p. 101): Extension to 3 events — 2 bags × 3 tiffin boxes × 2 bottles = 12 combinations.
• ROOT enumeration table (pp. 108–109): Side-by-side listing showing that treating the two O's as O1, O2 doubles each true arrangement, justifying division by 2! to give 4!/2! = 12.
• Fig. 6.3 (p. 115): Illustration of all three 2-player teams XY, YZ, ZX from {X, Y, Z}, showing that 3C2 = 3 (order does not matter).
• Process — choosing the right formula: ask three questions. (i) Does order matter? Yes ⇒ permutation; No ⇒ combination. (ii) Can repetitions occur? If yes, use n^r; if no, use nPr or nCr. (iii) Are any objects identical? If yes, divide by factorials of repeats.
• Process — counting word arrangements with restrictions: treat letters required to stay together as a single block, count arrangements of the reduced set, then multiply by the internal arrangements of the block (Examples 14, 16, pp. 112–113). For "no two X together" problems, place the other letters first and slot the X's into the gaps.
• Process — selection with constraints: split into cases by the constraint, compute each case separately, then sum. "At least k" usually means summing from k to maximum, or using total − "less than k".
• Process — at-least-one and at-most-one: complement frequently shortens the count. Number with "at least one" = total − number with "none".

2.4 Common confusions / NTA trap points

• Mixing up permutation vs combination — order matters → use nPr; order does not → use nCr. Word "arrangement"/"sequence"/"queue" signals P; words like "selection"/"committee"/"team"/"choose" signal C.
• Forgetting that 0! = 1 (definition, p. 105) — leads to wrong values of nP0 and nC0.
• Treating "with repetition" the same as "without repetition" — repetition allowed gives n^r (Theorem 2, p. 108), not nPr.
• Applying n! directly when some letters repeat — must divide by the factorials of repeated counts: ALLAHABAD = 9!/(4! 2!), not 9! (Example 9, p. 110).
• Forgetting the digit-0 restriction in number-formation problems — e.g., 3-digit numbers from {0,1,2,3,4,5} must exclude those with 0 in the hundreds place (Example 11, p. 110).
• In "vowels together" problems, students often forget to multiply by the internal arrangements of the vowel block (Example 14, p. 112; Example 16(ii), p. 113).
• Confusing nCa = nCb: it gives either a = b OR a + b = n (p. 117). Trap: students choose only one branch.
• Mis-counting "at least" by forgetting to subtract overlaps — e.g., "at least one boy AND at least one girl" requires careful complementation, not direct subtraction.
• Mistreating identical objects as distinct — adds spurious factors.
• Forgetting that nCr is symmetric: 10C7 = 10C3, so compute the smaller-r side for speed.
• Confusing the position of repetition: in "no letter is repeated" problems, both repetition and reuse are forbidden, so use nPr.
• Misreading "permutations of all n letters" as nPn = n! when in fact some letters repeat (then it is n!/p! …).
• Forgetting Pascal's identity is a binomial-coefficient identity, not a permutation identity.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 1, p. 102). Number of 4-letter words from ROSE without repetition?

Step 1 — 4 distinct letters, 4 places, no repetition: count = 4!. Step 2 — evaluate: 4! = 24. Step 3 — state: 24 words.

Example B (NCERT Example 4, p. 103). Number of signals from at least 2 flags out of 5 different flags arranged on a vertical staff?

Step 1 — split by number of flags used: 2, 3, 4, 5. Step 2 — compute each: 5P2 = 20; 5P3 = 60; 5P4 = 120; 5P5 = 120. Step 3 — sum: 20 + 60 + 120 + 120 = 320.

Example C (NCERT Example 9, p. 110). Permutations of ALLAHABAD?

Step 1 — count letters and repeats: 9 letters total; A repeats 4 times; L repeats 2 times. Step 2 — apply Theorem 4: 9!/(4! · 2!) = 362880/(24·2) = 362880/48. Step 3 — divide: = 7560.

Example D (NCERT Example 17, p. 117). If nC9 = nC8, find n.

Step 1 — apply equal-combinations rule: 9 ≠ 8, so use a + b = n. Step 2 — compute: n = 9 + 8 = 17. Step 3 — state: n = 17.

Example E (NCERT Example 18, p. 117). Committee of 3 from 2 men and 3 women: number with 1 man and 2 women?

Step 1 — count men selection: 2C1 = 2. Step 2 — count women selection: 3C2 = 3. Step 3 — multiply: 2 × 3 = 6 committees.