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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(n1)(n2)21n! = 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!(nr)!P(n,r)=\frac{n!}{(n-r)!}.
  • The number of combinations of rr objects chosen from nn distinct objects is (nr)=n!r!(nr)!\binom{n}{r}=\frac{n!}{r!(n-r)!}.
  • The symmetry identity is (nr)=(nnr)\binom{n}{r}=\binom{n}{n-r} because choosing rr items is equivalent to leaving out nrn-r items.
  • Pascal’s identity is (nr)=(n1r1)+(n1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r} for 1rn11\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)(nrk)=(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)ankbk(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!(nr)!P(n,r)=\frac{n!}{(n-r)!}.
Combination
A combination is an unordered selection, often counted by (nr)=n!r!(nr)!\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)=(n1r1)+(n1r)\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)ankbk(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!(nr)!\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 (n1r1)+(n1r)\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)ankbk\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)=(nnr)\binom{n}{r}=\binom{n}{n-r} makes sense using the idea of choosing items versus leaving items out.