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?
Show Solution $$P(8,3)=8(7)(6)=336$$
Final answer: 336 ways.
Committee Selection From 10 engineers, how many 4-person committees can be formed?
Show Solution $$C(10,4)=\frac{10!}{4!6!}=210$$
Final answer: 210 committees.
Arrangement with Repeated Letters How many distinct arrangements can be made from the letters in LEVEL?
Show Solution 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?
Show Solution 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?
Show Solution 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?
Show Solution 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?
Show Solution 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?
Show Solution 26 choices for each letter, 10 choices for each digit.
$$26^2\times10^3=676\times1000=676{,}000$$
Final answer: 676,000 passwords.
