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Combinatorial Identities Reference cheat sheet - grade 11-12

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Math Grade 11-12

Combinatorial Identities Reference Cheat Sheet

A printable reference covering factorials, permutations, combinations, binomial coefficients, Pascal’s identity, Vandermonde’s identity, and the binomial theorem for grades 11-12.

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Combinatorial identities help students count arrangements, selections, and algebraic patterns without listing every possibility. This cheat sheet covers the main formulas used in advanced algebra, precalculus, discrete math, and probability. Students need these identities to simplify expressions, solve counting problems, and recognize when different counting methods produce the same result.

The core ideas are factorials, permutations, combinations, and binomial coefficients. Important identities include symmetry, Pascal’s identity, the hockey-stick identity, Vandermonde’s identity, and the binomial theorem. Many formulas come from counting the same set in two different ways, which is a powerful strategy for proof and problem solving.

Key Facts

  • The factorial rule is n!=n(n−1)(n−2)⋯2⋅1n! = n(n-1)(n-2)\cdots 2\cdot 1 for positive integers, with 0!=10! = 1.
  • The number of permutations of rr objects chosen from nn distinct objects is P(n,r)=n!(n−r)!P(n,r)=\frac{n!}{(n-r)!}.
  • The number of combinations of rr objects chosen from nn distinct objects is (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}.
  • The symmetry identity is (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r} because choosing rr items is equivalent to leaving out n−rn-r items.
  • Pascal’s identity is (nr)=(n−1r−1)+(n−1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r} for 1≤r≤n−11\le r\le n-1.
  • The hockey-stick identity is ∑k=rn(kr)=(n+1r+1)\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}.
  • Vandermonde’s identity is ∑k=0r(mk)(nr−k)=(m+nr)\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}.
  • The binomial theorem is (a+b)n=∑k=0n(nk)an−kbk(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k.

Vocabulary

Factorial
A factorial n!n! is the product of all positive integers from 11 through nn, with 0!=10! = 1.
Permutation
A permutation is an ordered arrangement, often counted by P(n,r)=n!(n−r)!P(n,r)=\frac{n!}{(n-r)!}.
Combination
A combination is an unordered selection, often counted by (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}.
Binomial Coefficient
A binomial coefficient (nr)\binom{n}{r} counts the number of ways to choose rr objects from nn objects.
Pascal’s Identity
Pascal’s identity states that (nr)=(n−1r−1)+(n−1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.
Binomial Theorem
The binomial theorem expands powers using (a+b)n=∑k=0n(nk)an−kbk(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k.

Common Mistakes to Avoid

  • Using permutations when order does not matter is wrong because P(n,r)P(n,r) counts the same group multiple times in different orders.
  • Forgetting the factor r!r! in combinations is wrong because (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!} must remove the repeated orderings of each selected group.
  • Treating 0!0! as 00 is wrong because 0!=10! = 1, which makes formulas such as (n0)=1\binom{n}{0}=1 work correctly.
  • Applying Pascal’s identity with mismatched indices is wrong because (nr)\binom{n}{r} splits specifically into (n−1r−1)+(n−1r)\binom{n-1}{r-1}+\binom{n-1}{r}.
  • Expanding (a+b)n(a+b)^n without binomial coefficients is wrong because each term needs the multiplier (nk)\binom{n}{k} in (nk)an−kbk\binom{n}{k}a^{n-k}b^k.

Practice Questions

  1. 1 Compute (125)\binom{12}{5}.
  2. 2 How many ordered arrangements of 44 students can be chosen from a group of 1010 students?
  3. 3 Find the coefficient of x3x^3 in (2+x)7(2+x)^7.
  4. 4 Explain why (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r} makes sense using the idea of choosing items versus leaving items out.

Understanding Combinatorial Identities Reference

A useful way to understand factorials is to see them as a record of successive choices. If a team captain must assign distinct positions, the first position has many possible people, the next has one fewer, and the pattern continues. Multiplication works because each choice for one position can be paired with every valid choice for the next.

When a factorial fraction is simplified, most of its factors cancel. This is not just an algebra trick.

The canceled factors represent choices that do not matter after only part of an ordering is needed. The value of zero factorial keeps this logic consistent when no choices remain.

Many identities become clear when one item is treated differently from the rest. For Pascal’s identity, imagine forming a group of a fixed size from a larger set while focusing on one particular person. Every valid group either includes that person or excludes that person.

These two cases cannot overlap, yet together they include every possible group. This split into cases is one of the most reliable proof methods in combinatorics.

The hockey-stick pattern uses a related idea. It adds counts from rows of Pascal’s triangle, then interprets the total as choosing one larger group with a final element that marks where a smaller choice ended.

Vandermonde’s identity is important whenever a selection comes from two separate sources. Suppose a club has students from two grades and a committee must contain a fixed total number of members. One method chooses the committee all at once from the combined club.

Another method first decides how many members come from one grade, chooses those people, then fills the remaining places from the other grade. Adding every possible split must give the same total. The binomial theorem follows a similar choice process.

In each factor of a repeated product, one chooses either the first term or the second term. A coefficient tells how many different choice patterns create terms with the same powers.

These ideas appear in probability, computer science, genetics, card games, and statistical sampling. A probability calculation often starts by counting equally likely hands or groups. In algebra, expanding a power without multiplying every factor separately depends on recognizing the correct coefficient.

Students should first decide whether order changes the outcome. Choosing a president and a treasurer has order built into the roles. Choosing two people for an identical committee does not.

Students should state what each factor counts before multiplying or dividing. They should check boundary cases such as choosing nobody, choosing everybody, or choosing more objects than exist. Small examples with three or four objects often reveal an incorrect formula before it becomes a larger error.