Combination Identities Quiz: Practice Core Counting Formulas

The formula for combinations, written as ((n¦k)) = n!/(k!(n−k)!), works because we begin by counting all possible arrangements of k selected objects out of n using n!, which includes every ordering. Since combinations do not care about order, we divide by k! to remove the repeated orderings of the chosen items. We also divide by (n−k)! because the arrangement of items not chosen also creates unnecessary repetitions. After removing all overcounted arrangements, the result is the exact number of distinct, unordered subsets of size k.

To compute ((5¦2)), we use the combination formula. Factorial cancellation greatly simplifies the expression: 5!/(2!3!) becomes (5×4×3×2×1)/((2×1)(3×2×1)). After canceling identical factors in numerator and denominator, we are left with (5×4)/2. This equals 10, meaning there are exactly 10 different 2-element subsets that can be formed from a set of 5 items.

The identity ((n¦1)) = n follows directly from the combination formula. When selecting exactly one item out of n, each item is its own unique subset, so the result must be n. Algebraically, n!/(1!(n−1)!) simplifies because n! contains an (n−1)! factor, which cancels with the denominator, leaving only n.

The symmetry identity ((n¦k)) = ((n¦n−k)) reflects that selecting k items to include is equivalent to selecting n−k items to exclude. Thus, ((7¦4)) equals ((7¦3)), which is easier to compute. Evaluating 7!/(3!4!) gives 35. This symmetry is a core structural feature of Pascal’s triangle and appears in many combinatorial arguments.

The symmetry identity states that choosing 4 items to include is the same as choosing 3 items to leave out, so evaluating the smaller value (7 choose 3) makes the calculation easier and gives 35.

Because of the same symmetry property, ((10¦3)) and ((10¦7)) represent identical counts. Evaluating 10!/(3!7!) gives 120. This means there are 120 ways to either choose 3 items to include or choose 7 items to leave out from a set of 10. Symmetry helps reduce computational work by allowing us to use the smaller of k and n−k.

This identity is known as Pascal’s identity. It holds because any k-subset of an n-element set falls into exactly one of two categories: subsets that do not contain a specific element (counted by ((n−1¦k))) and subsets that do contain that element (counted by ((n−1¦k−1))). Combining these disjoint cases accounts for all possible k-subsets of the larger set.

The expression ((4¦0))+((4¦1))+((4¦2))+((4¦3))+((4¦4)) lists the entire fourth row of Pascal’s triangle. Adding these values gives 16, which equals 2⁴. This result follows from the fact that each element in a 4-item set can either be included or excluded in a subset, giving 2 choices per item and 2⁴ total subsets.

This identity states that adding every binomial coefficient in row n produces the total number of subsets of an n-element set. Each subset corresponds to a unique selection of items, and since each item can independently be chosen or not chosen, there are exactly 2ⁿ distinct subsets.

This identity describes how Pascal’s triangle is built. Each entry in row n+1 at position k+1 is formed by adding the two entries above it from row n. Combinatorially, it represents combining the subsets that include a new element with those that do not, explaining why the two corresponding entries merge into a single value.

By Pascal’s identity, adding ((5¦2)) and ((5¦3)) produces ((6¦3)) because these two coefficients sit directly above ((6¦3)) in Pascal’s triangle. This aligns with the combinatorial idea that subsets of size 3 from 6 items can be formed by considering whether the 6th item is included or not.

The hockey-stick identity says that a diagonal sum of binomial coefficients collapses into a single coefficient farther down the triangle. Beginning with ((3¦3)) and summing entries with constant lower index 3 produces ((7¦4)), illustrating how diagonals accumulate into a “handle” value.

The values ((2¦2)), ((3¦2)), ((4¦2)), ((5¦2)), ((6¦2)) produce 1, 3, 6, 10, and 15. These are triangular numbers, and their sum equals 35. The hockey-stick identity predicts that this diagonal sum equals ((7¦3)), showing consistency between numeric patterns and combinatorial identities.

A special identity states that the sum of squares of all binomial coefficients in row n equals the central coefficient of row 2n. For n = 3, summing ((3¦0))² + ((3¦1))² + ((3¦2))² + ((3¦3))² yields 20, which matches ((6¦3)). This identity reflects how products of subsets can relate to larger combinatorial structures.

When n is even, the sum of the even-index binomial coefficients equals 2ⁿ⁻¹. For n = 8, this gives 2⁷ = 128. This occurs because exactly half of all subsets of an 8-element set contain an even number of elements, due to symmetry in Pascal’s triangle.

This identity expresses the idea of choosing r items from two separate groups of sizes m and n. For each possible value of k, we choose k items from the first group and r−k items from the second group. Summing over all valid k gives the total number of ways to choose r items from all m+n items, represented by ((m+n¦r)).

The value ((n¦k)) equals zero whenever k > n because it is impossible to select more items than are available. This convention preserves the structure of binomial identities and ensures Pascal’s triangle behaves consistently at the edges.

There is exactly one empty subset and exactly n subsets containing only one element, so together they total n+1. These form the first two entries of row n in Pascal’s triangle and reflect the simplest, most basic combinatorial choices.

The identity ((n¦k)) = ((n−1¦k)) + ((n−1¦k−1)) is universally valid because any subset of size k either includes the last element or does not. This logic remains true regardless of which element is labeled last, making Pascal’s recurrence a fundamental combinatorial principle.

Binomial coefficients grow steadily as k approaches n/2 and then symmetrically decrease. This happens because there are generally more ways to form subsets of medium size than subsets that are very small or very large, so the coefficients peak at k = floor(n/2) or k = ceiling(n/2).