Tenth Grade Mathematics: Fundamental Counting Principle, Permutations, and Combinations

Counting discrete possibilities is one of the most practical and ancient branches of mathematics. While counting small collections can be achieved through brute-force enumeration, complex systems involving millions of combinations—such as digital passwords, genetic codon sequences, tournament schedules, and cryptographic keys—require systematic algebraic counting rules. Combinatorics provides this mathematical framework through two core principles: the Fundamental Counting Principle, which governs sequential choices, and the formal distinction between ordered arrangements (permutations) and unordered selections (combinations).

The Fundamental Counting Principle

The Fundamental Counting Principle (also termed the Multiplication Principle) governs multi-stage decision processes:

Theorem (The Multiplication Principle of Counting):
If an event can occur in m different ways, following which a second event can
occur in n different ways, then the total number of occurrences of the two
events in the specified order is:
                              m * n

Generalized to k Sequential Decisions:
If task 1 has n1 ways, task 2 has n2 ways, ..., task k has nk ways:
                    Total Ways = n1 * n2 * n3 * ... * nk

The Addition Principle:
If task A can be done in m ways and task B can be done in n ways, and the tasks
are MUTUALLY EXCLUSIVE (cannot both be chosen), then choosing either A OR B has:
                              m + n ways

The Factorial Notation

To streamline the algebra of combinatorics, we define the factorial of a non-negative integer n, denoted by n! (read as "n factorial"):

Formal Definition:
For any positive integer n:
               n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1

Special Axiomatic Value for Zero Factorial:
                             0! = 1
Why does 0! = 1?
Notice that n! = n * (n - 1)!  ===>  (n - 1)! = n! / n.
Setting n = 1: (1 - 1)! = 1! / 1  ===>  0! = 1 / 1 = 1.
This convention preserves the validity of permutation and combination formulas.

Permutations: Ordered Arrangements

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time:

KEY CONCEPT: ORDER MATTERS!
In permutations, the sequence of arrangement is essential:
(A, B, C) is DIFFERENT from (B, A, C) or (C, B, A).

Theorem (Permutation Formula nPr):
The number of permutations of n distinct objects taken r at a time (where 0 <= r <= n) is:
                                      n!
                       nPr = P(n, r) = ---------
                                      (n - r)!

Permutation of All n Objects (r = n):
The number of ways to arrange all n distinct objects in a line is:
                 nPn = n! / (n - n)! = n! / 0! = n!

Combinations: Unordered Selections

A combination is a grouping or selection of objects where the order of selection does not matter:

KEY CONCEPT: ORDER DOES NOT MATTER!
In combinations, choosing members for a team or committee:
{Alice, Bob} is IDENTICAL to {Bob, Alice}.

Theorem (Combination Formula nCr):
The number of combinations of n distinct objects taken r at a time (where 0 <= r <= n) is:
                                      n!
                    nCr = C(n, r) = ------------
                                     r! * (n - r)!

Fundamental Relationship Between nPr and nCr:
Each combination of r objects can be arranged internally in r! ways:
                        nPr = r! * nCr
                        nCr = nPr / r!

Symmetry Property of Combinations:
Choosing r objects to INCLUDE is mathematically identical to choosing
(n - r) objects to EXCLUDE:
                         nCr = nC(n - r)

Distinguishing Permutations from Combinations

+-----------------------+------------------------------------+-----------------------+
| Situation             | Does Order Matter?                 | Classification        |
+-----------------------+------------------------------------+-----------------------+
| Passwords, PIN codes  | YES (1234 != 4321)                 | Permutation (nPr)     |
+-----------------------+------------------------------------+-----------------------+
| Race finishes (1st,2nd| YES (Gold != Silver)               | Permutation (nPr)     |
+-----------------------+------------------------------------+-----------------------+
| Committee of people   | NO (Team {A,B} = Team {B,A})       | Combination (nCr)     |
+-----------------------+------------------------------------+-----------------------+
| Handshake problems    | NO (Handshake A-B = Handshake B-A) | Combination (nCr)     |
+-----------------------+------------------------------------+-----------------------+
| Lottery numbers drawn | NO (All chosen numbers win)        | Combination (nCr)     |
+-----------------------+------------------------------------+-----------------------+

Chapter Exercises

Exercise 1: Find the number of 4-letter words, with or without meaning, that can be formed out of the letters of the word 'ROSE', where the repetition of the letters is not allowed.

Exercise 2: Evaluate the following numerical values: (a) 7P3; (b) 10C3; and (c) verify that 12C5 = 12C7 using the symmetry property.

Exercise 3: A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Solutions and Explanations

Solution 1: The word 'ROSE' consists of 4 distinct letters: R, O, S, E. We wish to form 4-letter words without repetition, which corresponds to arranging all 4 distinct letters. Using the Fundamental Counting Principle: the 1st position can be filled by any of the 4 letters; the 2nd position can be filled by any of the remaining 3 letters; the 3rd position by any of the remaining 2 letters; and the 4th position by the 1 remaining letter. Total number of words = 4 * 3 * 2 * 1 = 4! = 24 words. Alternatively, using the permutation formula: 4P4 = 4! / (4 - 4)! = 24 / 0! = 24 / 1 = 24.

Solution 2: For part (a), 7P3 = 7! / (7 - 3)! = 7! / 4! = (7 * 6 * 5 * 4!) / 4! = 7 * 6 * 5 = 210. For part (b), 10C3 = 10! / [3! * (10 - 3)!] = 10! / (3! * 7!) = (10 * 9 * 8 * 7!) / [(3 * 2 * 1) * 7!] = (10 * 9 * 8) / 6 = 720 / 6 = 120. For part (c), by the combination symmetry property nCr = nC(n - r): with n = 12 and r = 5, n - r = 12 - 5 = 7. Thus: 12C5 = 12! / (5! * 7!) and 12C7 = 12! / (7! * 5!). Since the denominators are identical, 12C5 = 12C7 = (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1) = 792.

Solution 3: For the first question, a committee of 3 persons is to be chosen from a total group of 2 + 3 = 5 people. In a committee, the order of selection does not matter; hence, this is a combination problem. The total number of ways to choose 3 persons out of 5 is: Total Committees = 5C3 = 5! / (3! * 2!) = (5 * 4) / 2 = 10 ways. For the second question, we require the committee to consist of exactly 1 man and 2 women. The 1 man must be chosen from the 2 available men: 2C1 = 2 ways. The 2 women must be chosen from the 3 available women: 3C2 = 3! / (2! * 1!) = 3 ways. By the Fundamental Counting Principle, the total number of ways to form this specific committee is: 2C1 * 3C2 = 2 * 3 = 6 ways.