The most important concept in counting is whether order matters. This single distinction separates combinations from permutations and determines which formula to use.
Combinations (nCr): Order does not matter. Choosing which 3 friends from 10 to invite to a party is a combination — the group {Alice, Bob, Charlie} is the same regardless of what order you list them. Formula: C(n,r) = n! / (r!(n−r)!)
Permutations (nPr): Order matters. Arranging 3 friends in 3 specific seats (window, middle, aisle) is a permutation — Alice-window-Bob-middle-Charlie-aisle is different from Alice-aisle-Bob-window-Charlie-middle. Formula: P(n,r) = n! / (n−r)!
The relationship between them: P(n,r) = C(n,r) × r!. Every combination of r items can be arranged in r! different ways (permutations). So the number of permutations is always the number of combinations multiplied by r factorial.
The factorial of n (written n!) is the product of all positive integers from 1 up to n. So 5! = 5 × 4 × 3 × 2 × 1 = 120. By definition, 0! = 1, which is required for the combination formula to work correctly when r = 0 or r = n.
Factorials grow astonishingly fast:
This explosive growth is why large combinations seem astronomically improbable. The UK lottery has about 45 million possible combinations of 6 from 59 — which sounds large but is tiny compared to 52! arrangements of a card deck, which exceeds the estimated number of atoms in the observable universe (roughly 1080).
Why does C(n,r) = n! / (r!(n−r)!)? Here is the derivation. Start by counting the number of ways to arrange r items chosen from n in a sequence (permutations): P(n,r) = n × (n−1) × ... × (n−r+1) = n!/(n−r)!. But this counts each selection r! times (once for each ordering of the same r items). Dividing by r! removes the ordering and gives combinations: C(n,r) = P(n,r) / r! = n! / (r!(n−r)!).
Example: C(5,2) = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 20/2 = 10. Check: list the pairs from {A,B,C,D,E}: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE — exactly 10 pairs. ✓
Pascal's triangle is a triangular array where each number is the sum of the two numbers directly above it. Crucially, the entry in row n at position r (counting from 0) is exactly C(n,r):
Row 0: 1 = C(0,0)
Row 1: 1, 1 = C(1,0), C(1,1)
Row 2: 1, 2, 1 = C(2,0), C(2,1), C(2,2)
Row 3: 1, 3, 3, 1
Row 4: 1, 4, 6, 4, 1
This connects directly to the Binomial Theorem: (a+b)n = ∑r=0n C(n,r) an−r br. For example, (a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4. The coefficients 1, 4, 6, 4, 1 are exactly row 4 of Pascal's triangle. At A-Level (Year 2 Pure Maths), the Binomial Theorem is extended to non-integer exponents using the general binomial series.
Pascal's triangle contains many other patterns: the triangle's diagonal gives natural numbers, triangular numbers, and tetrahedral numbers. The row sums are powers of 2: row n sums to 2n, because ∑ C(n,r) = 2n (the total number of subsets of a set with n elements).
Combinations are the foundation of many probability calculations. When all outcomes are equally likely, P(event) = favourable outcomes / total outcomes. The total outcomes are often a combination count.
Card games: In a standard 52-card deck, the number of 5-card poker hands is C(52,5) = 2,598,960. The number of royal flushes (10, J, Q, K, A of same suit) is 4 (one per suit). So P(royal flush) = 4 / 2,598,960 ≈ 1 in 649,740.
Lottery: UK Lotto picks 6 from 59: C(59,6) = 45,057,474 possible combinations. The probability of a jackpot with one ticket is 1 in 45,057,474.
Birthday problem: How many people in a room before it is more likely than not that two share a birthday? Using permutations and combinations: with 23 people, the probability exceeds 50%. With 70 people, it exceeds 99.9%. This surprising result (the "birthday paradox") illustrates how quickly probabilities accumulate.
Combinations appear constantly in everyday decisions and organisation:
At A-Level Maths (Year 1 Pure), the Binomial expansion of (1+x)n for positive integer n is: (1+x)n = 1 + nCr(n,1)x + C(n,2)x2 + ... + xn. Students must be able to find specific terms, such as the coefficient of x3 in (2+3x)5. The general term is C(5,3) × 22 × (3x)3 = 10 × 4 × 27x3 = 1080x3.
At A-Level Year 2, the binomial series extends to any rational exponent n (including negatives and fractions), giving an infinite series valid for |x| < 1: (1+x)n = 1 + nx + n(n−1)x2/2! + ... where the coefficients are generalised combinations.
The Binomial distribution B(n, p) uses nCr at its core. If an experiment has probability p of success and is repeated n times independently, the probability of exactly k successes is: P(X=k) = C(n,k) × pk × (1−p)n−k. This is used in GCSE and A-Level statistics for problems like "probability of getting exactly 3 heads in 8 coin flips".
Example: Probability of exactly 2 heads in 5 fair coin flips: C(5,2) × (0.5)2 × (0.5)3 = 10 × 0.25 × 0.125 = 10 × 0.03125 = 0.3125 (31.25%).
Sometimes we allow the same item to be chosen more than once. The number of ways to choose r items from n types with repetition allowed is: H(n,r) = C(n+r−1, r). Example: choosing 3 ice cream scoops from 5 flavours (can repeat): H(5,3) = C(7,3) = 35. This is also called a "multiset coefficient" or "stars and bars" in combinatorics.
Combinations (nCr) count selections where order does not matter. Permutations (nPr) count arrangements where order does matter. Choosing 3 people for a committee from 10 is a combination (same group regardless of order). Arranging those 3 people in 3 specific roles is a permutation (different order = different assignment). nPr = nCr × r!, because each combination can be arranged in r! ways.
C(n,r) = n! / (r! × (n−r)!). A shortcut: only compute r factors in the numerator: n × (n−1) × ... × (n−r+1) and divide by r!. Example: C(10,3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120. Your scientific calculator has an nCr button (often labelled C or ³C on older models). In Excel: COMBIN(n,r).
A factorial (n!) is the product of all positive integers from 1 to n: n! = n × (n−1) × ... × 2 × 1. So 5! = 120, 10! = 3,628,800. By convention, 0! = 1 (required for combination formulas to work when r = 0 or r = n). Factorials grow extremely quickly: 20! exceeds 2 quadrillion.
This is a permutation: P(5,3) = 5! / (5−3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 120 / 2 = 60 ways. Alternatively: the first seat has 5 choices, the second seat has 4 remaining choices, the third seat has 3 remaining choices: 5 × 4 × 3 = 60.
The UK National Lottery (Lotto) requires choosing 6 numbers from 59. The total combinations are C(59,6) = 45,057,474. So the probability of winning the jackpot with a single ticket is 1 in 45,057,474 (approximately 1 in 45 million). Matching 5 balls (one of the secondary prizes) requires C(6,5) × C(53,1) = 6 × 53 = 318 ways, giving odds of about 1 in 144,415.
The entry in row n, position r of Pascal's triangle (counting from 0) is exactly C(n,r). So row 4 is C(4,0), C(4,1), C(4,2), C(4,3), C(4,4) = 1, 4, 6, 4, 1. These coefficients appear in the binomial expansion (a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4. The triangular structure follows Pascal's identity: C(n,r) = C(n−1,r−1) + C(n−1,r).