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Counting Techniques

Counting techniques answer: "How many ways can this happen?" The key question is: does order matter? Choosing 1st, 2nd, 3rd place (order matters → Permutation). Picking a 3-person team where roster order doesn't matter → Combination. The Fundamental Counting Principle says: if step 1 has $m$ choices and step 2 has $n$ choices, there are $m \times n$ total arrangements. For repeated items (e.g., letters in a word), divide by the factorial of each repeated count to avoid over-counting. Circular permutations (people sitting around a table) use $(n-1)!$ because one position is fixed as reference.

$$P(n,r)=\frac{n!}{(n-r)!} \quad \text{(order matters)}$$
$$C(n,r)=\frac{n!}{r!(n-r)!} \quad \text{(order does not matter)}$$

Permutation of Officers

From 8 students, how many ways can a president, vice president, and secretary be chosen?

$$P(8,3)=8(7)(6)=336$$

Final answer: 336 ways.

Committee Selection

From 10 engineers, how many 4-person committees can be formed?

$$C(10,4)=\frac{10!}{4!6!}=210$$

Final answer: 210 committees.

Arrangement with Repeated Letters

How many distinct arrangements can be made from the letters in LEVEL?

There are 5 letters, with E repeated twice and L repeated twice.

$$\frac{5!}{2!2!}=30$$

Final answer: 30 arrangements.

Fundamental Counting Principle

A restaurant offers 4 appetizers, 6 main courses, and 3 desserts. How many different 3-course meals are possible?

Multiply the number of choices at each step independently.

$$4\times6\times3=72$$

Final answer: 72 different meals.

Circular Permutation

In how many ways can 6 people be seated around a circular table?

For circular arrangements, one person is fixed as reference and the rest arrange around them. This removes identical rotations.

$$(6-1)!=5!=120$$

Final answer: 120 ways.

Mixed Group Selection

A committee of 5 must be chosen from 7 men and 4 women, with exactly 3 men and 2 women. How many ways are there?

Choose 3 men from 7 AND 2 women from 4 (independently).

$$C(7,3)\times C(4,2)=35\times6=210$$

Final answer: 210 ways.

Permutation with a Restriction

How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5 (no repetition) if the number must be even?

The last digit must be even: 2 or 4 (2 choices). The remaining 3 positions use any 3 of the remaining 4 digits.

$$2\times P(4,3)=2\times4\times3\times2=48$$

Final answer: 48 numbers.

Password Counting

A password consists of 2 letters (A–Z) followed by 3 digits (0–9). Letters and digits may repeat. How many passwords are possible?

26 choices for each letter, 10 choices for each digit.

$$26^2\times10^3=676\times1000=676{,}000$$

Final answer: 676,000 passwords.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w200

MSTE - Statistics and Probability / Fundamental Principles of Counting / MSTE May 2022

A music school produced a number of musicians which includes 3 drummers, 4 trumpet players and 5 pianists. How many different jazz trios can be formed from this batch of musicians if each trio consists of a drummer, a trumpet player and a pianist?

  1. 40
  2. 60
  3. 50
  4. 70
By the multiplication principle, $$N=n_{\text{drummer}}n_{\text{trumpet}}n_{\text{pianist}}=3(4)(5)=60.$$ Therefore, 60 different jazz trios can be formed.

Question Bank: w201

MSTE - Statistics and Probability / Fundamental Principles of Counting / MSTE May 2022

A company is designing fences using different widths of wooden boards. They have 5 boards that are 2 inches wide, 10 boards that are 3 inches wide, and 5 boards that are 4 inches wide. How many ways can the fence be designed using all these boards?

  1. 15,504
  2. 46,558,512
  3. 5,587,021,440
  4. 75,632,458
There are 20 boards, with repetitions of 5, 10, and 5 identical widths. The number of distinguishable arrangements is $$N=\frac{20!}{5!\,10!\,5!}=46{,}558{,}512.$$

Question Bank: w262

MSTE - Statistics and Probability / Probability / MSTE April 2025

Lots of 40 components each are called unacceptable if they contain 3 defectives or more. The procedure for sampling the lot is to select 5 components at random and reject the lot if a defective is found. What is the probability that exactly 1 defective is found in the sample if there are 3 defectives in the entire lot?

  1. 0.3011
  2. 0.2382
  3. 0.1668
  4. 0.1853
This is a hypergeometric problem: sampling without replacement from a lot of 40 containing $D = 3$ defectives and $G = 37$ good units.

Choose exactly 1 defective from the 3 and the remaining 4 from the 37 good units, out of all ways to choose 5 from 40:
$P = \dfrac{{}_{3}C_{1}\times{}_{37}C_{4}}{{}_{40}C_{5}}$
$P = \dfrac{3\times 66{,}045}{658{,}008}$
$\boxed{P = 0.3011}$

Question Bank: w266

MSTE - Statistics and Probability / Fundamental Principles of Counting / MSTE April 2025

10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?

  1. 144
  2. 131
  3. 115
  4. 90
Count every possible pair among the 17 people, then subtract the chairman-to-chairman pairs that never happen.

$N = {}_{17}C_{2} - {}_{7}C_{2}$
$N = 136 - 21$
$\boxed{N = 115}$

Question Bank: w285

MSTE - Statistics and Probability / Fundamental Principles of Counting / MSTE April 2025

In a certain municipality, all seven-digit telephone numbers begin with 321. How many telephone numbers may be assigned to that municipality if the last four digits should not begin or end in zero?

  1. 8,100
  2. 810,000
  3. 81,000
  4. 810
The first three digits are fixed as 3, 2, 1 (one choice each). Of the last four digits, the first and last exclude zero (9 choices each) while the middle two are free (10 choices each).

$N = 1\cdot 1\cdot 1\cdot 9\cdot 10\cdot 10\cdot 9$
$\boxed{N = 8{,}100}$

Question Bank: w298

MSTE - Statistics and Probability / Probability / MSTE April 2025

Four army recruits went to the supply room to get their military boots. Their shoe sizes were 7, 8, 9 & 10. The supply officer, after being informed of their sizes, prepared the four pairs of boots they need. If the boots are handed to each of the four recruits at random, what is the probability that none of them will receive the correct shoe size?

  1. 3/8
  2. 23/24
  3. 1/16
  4. 5/12
This is a derangement problem. Count the arrangements with exactly $k$ correct, then subtract.

Total number of ways $= 4! = 24$

Exactly 1 correct $= 1\cdot 2\cdot 1\cdot 1\cdot{}_{4}C_{1} = 8$
Exactly 2 correct $= 1\cdot 1\cdot 1\cdot 1\cdot{}_{4}C_{2} = 6$
Exactly 3 correct $= 0$  (fixing three forces the fourth)
Exactly 4 correct $= 1$

None correct $= 24 - 8 - 6 - 1 = 9$
$P = \dfrac{9}{24}$
$\boxed{P = \tfrac{3}{8}}$

Question Bank: w322

MSTE - Statistics and Probability / Fundamental Principles of Counting / MSTE April 2025

All members of fraternity play basketball while all but one play ice hockey; yet the number of possible basketball teams (5 members) is the same as the number of possible ice hockey teams (6 members). Assuming there are enough members to form either type of a team, how many are in the fraternity?

  1. 15
  2. 14
  3. 17
  4. 13
Let $n$ be the number in the fraternity. All $n$ play basketball, but only $n - 1$ play ice hockey.

Teams of basketball: ${}_{n}C_{5}$
Teams of hockey: ${}_{n-1}C_{6}$

Set them equal:
${}_{n}C_{5} = {}_{n-1}C_{6}$
$\dfrac{n!}{(n-5)!\,5!} = \dfrac{(n-1)!}{(n-7)!\,6!}$

Cancelling the common factors leaves
$\dfrac{n}{5!} = \dfrac{(n-5)(n-6)}{6!}$
$6n = (n-5)(n-6)$
$6n = n^{2} - 11n + 30$
$n^{2} - 17n + 30 = 0$

Roots: $n_1 = 15$ and $n_2 = 2$. Discard $n = 2$ (too few to form a team).
$\boxed{n = 15}$