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Grade 11-12 (age 16-18)

Combination

Combination

A combination is a selection where you just pick, without caring about the order. Where a permutation told first, second, and third apart, a combination only asks who got chosen, so you divide out the r! orderings that overlap and get nCr = nPr / r!. Stack these values into a triangle and Pascal's triangle appears, where each entry is the sum of the two above it, and the left-right symmetry nCr = nC(n-r) is plain to see. Use the total-count n and choose-count r sliders to watch the picks gather into one unordered group, and grow the triangle with the rows slider to check the pattern.

Order-free Selection
🧩 What is a Combination?
①Pick r out of n without regard to order
②(A, B, C) and (C, A, B) are the same combination
③Divide a permutation by r! to get the combination
Visualize nCr
5
3
Combination Formula
nCr = n!r!(n-r)! = nPrr!
Number of ways to pick r from n (no order)
💡 Permutation → Combination
①Permutation (nPr) = pick + arrange
②Combination (nCr) = pick only
③nPr = nCr × r! → nCr = nPr ÷ r!
④Remove duplicate orderings to convert P → C
Pascal's Triangle
6
Pascal's Identity
nCr = n-1Cr-1 + n-1Cr
Sum of two upper entries = entry below
Properties of Combinations
Symmetry
nCr = nCn-r
Choosing r = leaving out (n−r)
Sum of Combinations
nC0 + nC1 + ··· + nCn = 2n
Total number of subsets of n elements
Work It Out
Example 1
Find ₆C₂, the number of ways to choose 2 of 6 distinct items.
1
A combination divides the permutation by the factorial of the number chosen.
₆C₂ = (6 × 5)/(2 × 1)
2
Compute.
= 30/2 = 15
15
A combination ignores order, so divide the permutation by r!.
Example 2
From 4 boys and 3 girls, choose 3 people with exactly 1 girl. How many ways?
1
Choose 1 girl as ₃C₁ and 2 boys as ₄C₂, then multiply.
₃C₁ × ₄C₂
2
Compute.
= 3 × 6 = 18
18
With a condition, choose group by group and multiply (product rule).
Wrap-up
Core Formula
nCr = n!r!(n-r)!
Pick r without order
Grade-11 school exam type
How many ways are there to pick 2 representatives from 8 people?
16
21
28
36
56
③ 28
1
Choosing 2 without regard to order is ₈C₂.
₈C₂ = (8 × 7)/(2 × 1)
2
Compute.
= 56/2 = 28
🎯 Exam Points
①P vs C: order matters?
②nCr = nPr ÷ r!
③Pascal: sum of upper two
④Symmetry: nCr = nC(n−r)
⑤Sum of all nCk = 2^n
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Permutation
Next →
Binomial Theorem
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