Permutation, Combination & Probability

A permutation counts arrangements where order matters. A combination counts selections where order does not matter. Probability compares favourable outcomes with all equally likely outcomes.

Use the complement when "at least one" is easier to count as none. For sampling without replacement, update both numerator and denominator.

Choose one of three identical covered cards labelled A, B and C at random. What is the probability of B? 1. There are 3 equally likely choices. 2. Only one choice is B. 3. Probability = 1 ÷ 3 = 1/3.

The three cards have equal chances of being selected. Only B satisfies the requested event, giving probability 1/3. If the selection process favoured a card, counting cards alone would not determine the probability. In counting problems, distinguish an unordered selection from assigning different roles.

A bag has 5 red and 3 blue counters. Choose 3 without replacement. Probability of at least one blue=1-P(all red)=1-[5C3/8C3]=1-10/56=23/28. Ordered calculation gives the same all-red chance: (5/8)(4/7)(3/6)=5/28. There is no need to separately count one, two and three blues.

The equally likely condition matters. Picking two people is different from assigning them two different jobs. Independence cannot be assumed without replacement. Equally likely counters do not imply equally likely colour combinations.

A bag has 3 red counters and 2 blue counters. Each counter is equally likely to be picked. There are 5 possible counters and 3 are red, so the probability of red is 3/5. Picking a captain and a deputy is ordered; choosing two team members is not.

Do not assume every named category is equally likely. Red and blue are two colours, but their chances are not 1/2 here because the numbers of counters differ.

Use permutations when order changes the outcome and combinations when only membership matters. State whether repetition is allowed.

Choose two representatives from five people: 5C2 = 10. Choose a captain and deputy: 5 x 4 = 20, since exchanging their roles creates a different assignment.

Use the complement when "at least one" is easier to count as none. For sampling without replacement, update both numerator and denominator.

A bag has 5 red and 3 blue counters. Choose 3 without replacement. Probability of at least one blue=1-P(all red)=1-[5C3/8C3]=1-10/56=23/28. Ordered calculation gives the same all-red chance: (5/8)(4/7)(3/6)=5/28. There is no need to separately count one, two and three blues.

Independence cannot be assumed without replacement. Equally likely counters do not imply equally likely colour combinations.

Use the product rule for successive choices and the sum rule for disjoint alternatives. An arrangement cares about order; a selection does not. State whether repetition is allowed and whether objects and recipients are distinct. Key formulas For integers 0≤r≤n: nPr=n!/(n−r)! and nCr=n!/[r!(n−r)!]. Also nPr=r!×nCr, nCr=nC(n−r), and the sum of nCr over r=0,…,n is 2^n. Arrangements of n objects with repeated groups of sizes p,q,… number n!/(p!q!…). Distributing n identical objects among r named recipients with zero allowed gives C(n+r−1,r−1); with everyone receiving at least one, C(n−1,r−1), provided n≥r.

Choosing two representatives from five people gives C(5,2)=10. Assigning a captain and deputy gives P(5,2)=20. A bag with three red and two blue counters gives probability (3/5)(2/4)=3/10 of two reds without replacement. Probability rules For equally likely outcomes, P(E)=favourable/total. In general P(A∪B)=P(A)+P(B)−P(A∩B), and P(A∩B)=P(A)P(B|A) when P(A)>0. Independence allows P(B|A)=P(B). Mutually exclusive events have zero overlap and are not independent when both have positive probability.

Do not assume equally likely named categories when their underlying outcomes have different counts. “Without replacement” generally changes later probabilities.

Worked example 1 Four-digit numbers use only 1,2,3, with both 2 and 3 occurring at least once. Inclusion–exclusion gives 3⁴−2⁴−2⁴+1⁴=50. No leading-zero restriction is needed because zero is absent. Worked example 2 Distribute 20 identical balloons to four named children, with each receiving a positive even number. Divide every allocation by 2: four positive integers must sum to 10. The number of allocations is C(9,3)=84 by placing three separators among nine gaps. Worked example 3 In a 75-question test, a correct answer earns 3, a wrong answer loses 1 and an unanswered question earns 1. Let c,w,u be their counts. Then c+w+u=75 and 3c−w+u=97, giving c−w=11. If unanswered questions exceed attempted questions, c+w≤37. Thus 2w+11≤37, so w≤13 and c≤24. Taking c=24,w=13,u=38 attains the bound, proving that 24 is the maximum. Worked example 4 A well-mixed 40 litre milk container has 4 litres removed and replaced with water each time. The milk fraction after n replacements is 0.9^n. Since 0.9⁶=0.531441 and 0.9⁷=0.4782969, seven replacements are the first to leave less milk than water.

The marking rule in the third example belongs to that hypothetical problem, not a claim about an official exam. A maximum requires both a bound and a valid case achieving it.