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Lesson 5 of 8

Probability · Lesson 5 of 8

Counting Arrangements and Selections

“Choose permutations or combinations to count larger sample spaces and their favourable outcomes.”

Learning Objectives

• Decide whether an outcome is an ordered arrangement or an unordered selection. • Use multiplication of choice counts, factorials, permutations, and combinations consistently. • Count position, precedence, and adjacency restrictions in random orders. • Find probabilities for committees, card hands, and race finishers. • Account for repetition restrictions and strict numerical bounds.

Counting replaces a long list

A few coin tosses or two dice can be listed directly. A seven-card hand cannot be handled comfortably that way: there are many millions of possible hands. The probability principle does not change. We still compare a favourable count with a total count, but we use counting arguments to obtain those numbers.

Start by deciding what one outcome records. Does a committee record only its members, or also a chairperson? Does a race record the order of finishers? Does a card hand record the order of drawing? Counting must match that description in both the numerator and the denominator. A correct count of the wrong kind of outcome still gives a wrong model.

The multiplication principle counts stages of a construction. If each of a first set of choices allows k second choices, the number of complete two-stage choices is the first count multiplied by k. This is a statement about counting; it does not assume a probability multiplication rule. For arrangements without repetition, the available count drops after each choice.

Definition
Factorial

For a positive integer n, n!=n(n−1)⋯2·1. Also 0!=1. Factorials count arrangements of distinct objects.

Ordered selections without repetitionLaTeX
Choose r distinct items from n and record their order. There are n choices for the first position, n−1 for the second, and so on.

For an unordered selection, each chosen group is counted r! times by the ordered selection count, once for every order of its r members. Dividing by r! removes that repetition. It does not remove actual different groups; it removes repeated descriptions of the same group.

Unordered selections without repetitionLaTeX
Choose r distinct items from n with no role assigned to their order. The condition is 0≤r≤n.
Common mistake

Use the same outcome convention in both counts. For a two-person committee, {A,B} and {B,A} are one group. For first and second places in a race, AB and BA are different results.

Choosing a committee

A committee problem normally treats every group of the stated size as one outcome. When the problem says the group is chosen uniformly, those groups have equal probability. Membership restrictions can often be counted by choosing separately from the relevant categories.

Example — Two members from two men and two women

Problem
Select a two-person committee uniformly from two men and two women. Find the probabilities of no man, one man, and two men.

  1. 1.There are four people. The total number of unordered committees is C(4,2)=4×3/(2×1)=6.
  2. 2.No man means both women: C(2,2)=1 committee, so its probability is 1/6.
  3. 3.One man means one woman as well. Choose the man in C(2,1)=2 ways and the woman in C(2,1)=2 ways: 2×2=4 committees, giving 4/6=2/3.
  4. 4.Two men gives C(2,2)=1 committee and probability 1/6.
  5. 5.The three categories are disjoint and exhaustive. Their probabilities add to 1/6+2/3+1/6=1, which checks the counts.
Committee typeNumber of committeesProbability
Two women11/6
One man and one woman42/3
Two men11/6

Restrictions on a random visiting order

If four distinct cities are visited in a random order with all complete orders equally likely, there are 4!=24 elementary outcomes. Different restrictions require different counting arguments. Before means earlier somewhere in the order, while just before means immediately adjacent on the left.

First citySix possible full orders
AABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BBACD, BADC, BCAD, BCDA, BDAC, BDCA
CCABD, CADB, CBAD, CBDA, CDAB, CDBA
DDABC, DACB, DBAC, DBCA, DCAB, DCBA
Example — Five visiting-order conditions

Problem
For a uniformly random order of A,B,C,D, find the chances of: A before B; A before B before C; A first and B last; A first or second; A immediately before B.

  1. 1.The total count is 24. Swapping A and B pairs every order with A before B with one having B before A. Each group has 12 orders, giving 1/2.
  2. 2.For A before B before C, the only relative order of those three letters is ABC. Insert D in one of four places: DABC, ADBC, ABDC, ABCD. The probability is 4/24=1/6.
  3. 3.A first and B last leaves the two middle positions for C,D in 2!=2 orders, giving 2/24=1/12.
  4. 4.A first allows 3!=6 orders, as does A second. The two cases cannot both occur, so the favourable count is 6+6=12 and the probability is 1/2.
  5. 5.A immediately before B lets us treat AB as one block. Arrange the three units AB,C,D in 3!=6 ways, giving 6/24=1/4. There is no factor of 2 inside the block, because BA would violate the condition.
Immediately before: keep AB togetherABCDThree units can be arranged in 3! = 6 ways.A before B also allows gaps: A C D B.
Adjacency is a block-counting condition— Joining AB works for immediately before. It would wrongly exclude valid orders if the question only said before.

Hands containing a specified number of kings

A hand is an unordered selection of cards. Uniform drawing without replacement from a well-shuffled deck gives equally likely seven-card sets. Counting exact numbers of kings requires selecting the kings and the remaining non-kings separately; those choices determine a unique hand.

Example — All, exactly three, and at least three kings

Problem
A seven-card hand is drawn uniformly from 52 cards. Find the probabilities of all four kings, exactly three kings, and at least three kings.

  1. 1.The total number of hands is C(52,7). There are 4 kings and 48 non-kings.
  2. 2.All four kings means selecting C(4,4)=1 king set and C(48,3) additional cards. The probability is C(4,4)C(48,3)/C(52,7)=1/7735.
  3. 3.Exactly three kings means choosing C(4,3) kings and C(48,4) non-kings. Its probability is C(4,3)C(48,4)/C(52,7)=9/1547.
  4. 4.At least three kings means exactly three or exactly four. These cases are disjoint, so add their probabilities.
  5. 5.Use 9/1547=45/7735. The required sum is (45+1)/7735=46/7735.
Counting exact king totalsLaTeX
The first factor chooses k kings; the second chooses the remaining 7−k non-kings. This is a counting argument for this seven-card question.

The non-king requirement is essential in the exact-three case. Choosing three kings and four cards from the whole remaining deck could include the fourth king, so it would count some all-four hands. Restricting those four cards to the 48 non-kings makes the cases precise.

Recording only the positions that matter

A sample space can record a complete order or only an initial part of it. If we use a shorter description, we must justify that the shorter outcomes have equal weight. In a uniform five-team finishing-order model, every ordered top-three result has two possible orders for the remaining teams, so every top-three result has the same weight.

Example — The first three teams

Problem
Five teams A,B,C,D,E have equally likely finishing orders. Find the chance that A,B,C finish first, second, third respectively, and the chance that they fill those three places in any order.

  1. 1.There are P(5,3)=5×4×3=60 ordered top-three results. Each can be completed in 2! ways, so the 60 results are equally likely.
  2. 2.For the exact order ABC, there is one favourable top-three result, giving 1/60.
  3. 3.For any order of A,B,C, there are 3!=6 favourable results, giving 6/60=1/10.
  4. 4.Alternatively, use all 5!=120 complete orders. The exact order ABC has 2!=2 completions, while any order has 3!2!=12. These give the same fractions, 2/120 and 12/120.

Grouping many sequences by a numerical result

Some questions ask for the chance of a numerical quantity produced by a sequence. The quantity itself may have fewer possible values than the sequence has outcomes. Group the elementary outcomes according to their value, then add their equally likely weights within each group.

Example — Money after four tosses

Problem
Four fair tosses give a gain of ₹1 per head and a loss of ₹1.50 per tail. Find every possible total and its probability.

  1. 1.If h heads occur, there are 4−h tails. The total is h−1.5(4−h)=2.5h−6 rupees.
  2. 2.For h=0,1,2,3,4 the amounts are −6, −3.50, −1, 1.50, and 4. There are five distinct amounts.
  3. 3.There are 16 equally likely sequences. The sequence counts for these head totals are C(4,0),C(4,1),C(4,2),C(4,3),C(4,4)=1,4,6,4,1.
  4. 4.The probabilities of the five amounts, in the same order, are 1/16,4/16,6/16,4/16,1/16. The amounts themselves are not equally likely.
Number of heads01234
Amount (₹)−6−3.50−11.504
Probability1/161/43/81/41/16

Respecting every restriction

Restricted digit questions often have a small exceptional case that changes the count. Read the bound literally, decide whether repetition is allowed, and check whether zero is permitted in the first position. A number and a four-digit code are different objects: a code can begin with zero if the lock allows it.

Example — Strictly greater than 5,000

Problem
Form a four-digit number uniformly from digits 0,1,3,5,7, subject to being greater than 5,000. Find the probability of divisibility by 5 with repetition allowed, and then without repetition.

  1. 1.With repetition, the first digit can be 5 or 7. This initially gives 2×5³=250 numbers, but 5000 equals the bound and must be excluded. Therefore n(S)=249.
  2. 2.Divisibility by 5 requires a final digit 0 or 5. The initial favourable count is 2×5²×2=100. It also includes 5000, so the correct favourable count is 99.
  3. 3.The first probability is 99/249=33/83. A value of 2/5 would result from keeping 5000 in both counts, which answers a different bound.
  4. 4.Without repetition, either first digit leaves P(4,3)=24 completions, so there are 48 numbers. All exceed 5000 because three distinct following digits cannot all be zero.
  5. 5.If the first digit is 5, the last must be 0; the middle places have 3×2=6 orders. If the first digit is 7, the last may be 0 or 5, giving 6+6=12 orders.
  6. 6.Thus the no-repetition favourable count is 18, and its probability is 18/48=3/8.
Example — One specified code among admissible guesses

Problem
A four-wheel lock uses digits 0–9 and its code has no repeated digit. A person guesses uniformly among codes with no repetition. Find the probability of the correct guess.

  1. 1.The first wheel has ten choices, including 0; the next wheels have 9,8,7 choices because repetition is forbidden.
  2. 2.There are 10×9×8×7=5040 admissible sequences.
  3. 3.Exactly one sequence is correct, so the probability is 1/5040.
  4. 4.This answer assumes the person uses the known no-repetition rule. A uniform guess among all 10⁴ unrestricted sequences would use a different sample space and give 1/10000.

Quiz

Quick check

How many unordered two-person committees can be formed from four people?

Quick check

In a random order of four cities, what is the probability that A is immediately before B?

Quick check

Which numerator counts seven-card hands with exactly three kings?

Quick check

Why divide P(n,r) by r! for unordered groups?

Quick check

What is the chance that A,B,C fill the top three places of five teams in any order?

Quick check

With repeated digits allowed, why are there 249 rather than 250 allowed numbers strictly greater than 5000?

Practice Problems

Practice Problems
  1. For four fair tosses, gains are ₹1 per head and losses ₹1.50 per tail. List all possible total amounts and calculate the probability of each using the sixteen sequences or their head-count groups.
  2. A lottery selects one fixed set of six distinct numbers from 1 to 20. A player chooses a six-number set uniformly. Find the probability of matching it exactly; order does not matter.
  3. Four cards are drawn uniformly without replacement from 52 cards. Find the probability of exactly three diamonds and one spade. Explain the two factors in the numerator.
  4. Using digits 0,1,3,5,7, choose uniformly from four-digit numbers strictly greater than 5000. Find the chance of divisibility by 5 with repetition and without repetition. Treat 5000 correctly.
  5. A four-wheel lock uses a code of four distinct digits from 0–9. Find the chance of a correct uniform guess among admissible codes, including those beginning with zero.
  6. For a random order of A,B,C,D, count orders with A before B but not immediately before B. Explain how to subtract the smaller event from the larger one.
  7. Use full five-team orders and ordered top-three results to calculate the relay probabilities in two ways. Explain why both equally likely descriptions give the same answers.

Key Takeaways

Key Takeaways

• Choose an outcome convention before using a counting formula. • Permutations record order; combinations remove the r! orders of each selected group. • Multiply successive choice counts when each stage has the stated number of options, and separate disjoint cases carefully. • Before and immediately before impose different restrictions. • Strict bounds, repetition rules, and the distinction between numbers and codes can change both counts.

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Equally Likely Outcomes

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The Addition Rule for Events