Permutations, arrangements, combinations
Counting choices: lists with repetition, arrangements, permutations, anagrams, combinations, binomial coefficients, Pascal's triangle and the binomial theorem.
- 27 cards
- 5 topics
- Free
The outline, topic by topic
Each topic is weighted the way the exam weighs it.
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A few cards from the deck
Factorial n!: what is it?
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The product of the integers from 1 to n, with the convention 0! = 1; it is the number of ways to order n distinct objects.
6 talks can be given in 6! = 720 orders: 6 choices for the first, 5 for the second, and so on. 0! = 1 because there is exactly one way to order nothing.
How many passwords of 4 distinct letters can be formed from the 26 letters of the alphabet?
- 358 800
- 456 976
- 14 950
- 104
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358 800
Order matters and letters are distinct: 26 × 25 × 24 × 23 = 358,800. 456,976 = 26⁴ allows repetitions, and 14,950 = C(26, 4) ignores order.
Choosing p elements among n with order and with repetition allowed gives … lists; without repetition, you get … arrangements.
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Choosing p elements among n with order and with repetition allowed gives nᵖ lists; without repetition, you get n! / (n − p)! arrangements.
A 4-digit code: 10⁴ = 10,000 lists. A podium among 8: 8 × 7 × 6 = 336. At each rank, you keep n choices with repetition, one fewer without.
Arrangement of p out of n (p-permutation): what is it?
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An ordered list of p distinct elements taken from n; their number is n! / (n − p)! = n × (n − 1) × … × (n − p + 1).
A podium among 8 runners: 8 × 7 × 6 = 336. Order matters (winning is not finishing third) and a runner cannot take two places.