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Grade 11-12 (age 16-18)

high school Combination

A combination is a choice where only what you picked matters, not the order. A permutation of A, B and C also counts the order they are lined up in, so the same three items are counted 3! = 6 times as different cases; dividing the number of permutations by r! therefore gives the number of combinations. Stack these values row by row and you get Pascal's triangle, where the two numbers above add up to the one below, and its left-right symmetry shows ₙCᵣ = ₙCₙ₋ᵣ at a glance. Move the sliders for the total n and the number chosen r to watch the many orders of the chosen balls collapse into one combination, with the counts beside them; the rows slider extends the triangle downward.

Choosing Without Order

🧩 What a combination is
①Choosing r of n distinct items without regard to order is a combination, and the number of them is written ₙCᵣ
②Picking A, B, C or C, A, B gives the same combination
③Picking two representatives is a combination; picking a president and a vice president is a permutation, because the two posts differ

From Permutations to Combinations

5
3
Number of combinations
nCr = n!r!(n-r)! = nPrr!
The number of ways to choose r of n items without order
💡 Why divide by r!
①The permutation count ₙPᵣ counts the choice and the order it is lined up in
②The r chosen items can be lined up in r! ways, so each combination is counted r! times among the permutations
③So ₙPᵣ = ₙCᵣ × r!, which gives ₙCᵣ = ₙPᵣ ÷ r!
④In the figure, every order of the chosen balls gathers into a single bracketed group

Pascal's Triangle

6
Property of Pascal's triangle
nCr = n-1Cr-1 + n-1Cr
Two neighbors in one row add up to the number just below them

Properties of Combinations

Symmetry
nCr = nCn-r
Choosing r items is the same as choosing the n − r items to leave out
Sum of combinations
nC0 + nC1 + ··· + nCn = 2n
The number of subsets of an n-element set. Each element is either in or out, so the total is 2ⁿ

Worked Examples

Example 1
Find ₆C₂, the number of ways to choose 2 of 6 distinct items.
1
Divide ₆P₂ by 2!.
₆C₂ = (6 × 5)/(2 × 1)
2
Compute the numerator and the denominator.
= 30/2 = 15
▸ 15
6 × 5 = 30 counts the order too; dividing by the 2 orders of two items removes the order and leaves 15.
Example 2
Three people are chosen from 4 boys and 3 girls. In how many ways can exactly 1 girl be included?
1
Choose the 1 girl in ₃C₁ ways and the 2 boys in ₄C₂ ways; both choices happen together, so multiply.
₃C₁ × ₄C₂
2
Compute each number of combinations and multiply.
= 3 × 6 = 18
▸ 18
When there is a condition, choose from each group separately and multiply by the multiplication principle.

Summary

Number of combinations
nCr = n!r!(n-r)!
The number of ways to choose r items without order
Grade-11 school exam type
In how many ways can 2 representatives be chosen from 8 people?
①16
②21
③28
④36
⑤56
▸ ③ 28
1
The 2 people are chosen without order, so the count is ₈C₂.
₈C₂ = (8 × 7)/(2 × 1)
2
Compute the numerator and the denominator.
= 56/2 = 28
🎯 Exam Points
①What separates permutations from combinations is whether the order changes the outcome
②ₙCᵣ = ₙPᵣ ÷ r!
③In Pascal's triangle, the two numbers above add up to the one below
④Since ₙCᵣ = ₙCₙ₋ᵣ, when r is more than half of n it is easier to compute with n − r
⑤Adding ₙC₀ through ₙCₙ gives 2ⁿ
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Permutation
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Binomial Theorem

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