The Fundamental Principle of Counting

Counting Without Listing

How many outfits can Mohan make from 3 pants and 2 shirts? You could list all six — or notice the structure: 3 ways to pick a pant, and for every such choice, 2 ways to pick a shirt. So $3 \times 2 = 6$, no listing needed.

Key Point (Fundamental Principle of Counting): "If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, then the total number of occurrence of the events in the given order is $m \times n$." Also called the multiplication principle.

The principle chains to any number of events. Sabnam's 2 school bags, 3 tiffin boxes and 2 water bottles give $2 \times 3 \times 2 = 12$ ways of carrying one of each:

Multiply or add? Multiply when choices happen in succession (this AND then that). Add when the possibilities are alternatives that cannot happen together (this OR that) — as in the flag-signal problem, where a signal uses 2 or 3 or 4 or 5 flags and the four separate counts add up.

The Vacant-Places Method

Most counting problems in this chapter are solved by one picture: draw a box for each position, write in how many choices remain for that box, and multiply.

Model problem: 4-letter words from ROSE with no repetition — the first place has 4 choices, then 3, then 2, then 1: $4 \times 3 \times 2 \times 1 = 24$ words. With repetition allowed, every place keeps all 4 choices: $4^4 = 256$.

Key Point: Fill the most restricted place first. For 2-digit even numbers from 1-5 with repetition, start at the units place — only 2 or 4 work (2 ways) — then the tens place takes any of the 5 digits: $2 \times 5 = 10$ numbers.

[Board Tip] Read the repetition clause twice. "No digit repeated" shrinks the choices box by box ($5, 4, 3, \ldots$); "digits can be repeated" keeps every box full. Half the errors in this topic are repetition misreads.

[JEE Tip] When a restriction involves a specific position (last digit even, first letter a vowel, number starts with 67), lock those positions first and count the free positions afterwards — the multiplication principle does not care in which order you fill the boxes, only that you fix one order and count consistently.

Solved Examples

Example 1: Two-flag signals

Given 4 flags of different colours, how many signals can be generated if a signal requires 2 flags, one below the other?

Solution:

Step 1 — Fill the upper place. Any of the 4 flags: 4 ways.

Step 2 — Fill the lower place. Any of the remaining 3 flags: 3 ways.

Step 3 — Multiply. $4 \times 3 = 12$ signals — order matters: red over blue is a different signal from blue over red.

Takeaway: Successive choices multiply; the ordering of positions is what makes this a signal count, not a selection count.

Example 2: Even numbers with repetition

How many 2-digit even numbers can be formed from 1, 2, 3, 4, 5 if digits can be repeated?

Solution:

Step 1 — Fill the most restricted place first. Units must be even: 2 or 4 — 2 ways.

Step 2 — Fill the tens place. Repetition allowed, so any of the 5 digits: 5 ways.

Step 3 — Multiply. $2 \times 5 = 10$ numbers.

Takeaway: Lock the restricted position first; the free positions follow.

Example 3: Add the alternatives

Five different flags are available. How many signals can be generated by arranging at least 2 flags in order on a vertical staff?

Solution:

Step 1 — Split into mutually exclusive cases. A signal uses exactly 2, 3, 4 or 5 flags — no overlap between cases.

Step 2 — Count each case by the multiplication principle. 2 flags: $5 \times 4 = 20$; 3 flags: $5 \times 4 \times 3 = 60$; 4 flags: $5 \times 4 \times 3 \times 2 = 120$; 5 flags: $5 \times 4 \times 3 \times 2 \times 1 = 120$.

Step 3 — Add across cases. $20 + 60 + 120 + 120 = 320$ signals.

Takeaway: Multiply WITHIN a case (successive choices), add ACROSS cases (alternatives).

Example 4: Repetition on vs off

How many 3-digit numbers can be formed from 1, 2, 3, 4, 5 if (i) repetition is allowed (ii) repetition is not allowed?

Solution:

Step 1 — (i) Repetition allowed. Every place keeps all 5 choices: $5 \times 5 \times 5 = 125$.

Step 2 — (ii) Repetition off. Choices shrink place by place: $5 \times 4 \times 3 = 60$.

Takeaway: The repetition clause decides whether the boxes stay full or shrink — read it twice.

Example 5: Even with repetition, six digits

How many 3-digit even numbers can be formed from 1, 2, 3, 4, 5, 6 if digits can be repeated?

Solution:

Step 1 — Units place first. Even digits available: 2, 4, 6 — 3 ways.

Step 2 — Free places. Tens and hundreds each take any of the 6 digits: $6 \times 6$.

Step 3 — Multiply. $6 \times 6 \times 3 = 108$.

Takeaway: With repetition on, only the restricted place differs from a free place.

Example 6: Letter codes

How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter repeats?

Solution:

Step 1 — Set up the vacant places. Four boxes with shrinking choices: $10, 9, 8, 7$.

Step 2 — Multiply. $10 \times 9 \times 8 \times 7 = 5040$ codes.

Takeaway: A no-repetition code count is a falling product — the start of a factorial.

Example 7: Telephone numbers

How many 5-digit telephone numbers can be constructed using digits 0-9 if each number starts with 67 and no digit appears more than once?

Solution:

Step 1 — Lock the fixed positions. The first two places are 6, 7 — both now used up.

Step 2 — Count the free places. Three places choose from the 8 unused digits: $8 \times 7 \times 6$.

Step 3 — Multiply. $336$ numbers.

Takeaway: Locked positions consume their digits — subtract them from the pool before counting the free places.

Example 8: Tosses and flags

(i) A coin is tossed 3 times; how many outcome sequences are possible? (ii) With 5 flags of different colours, how many 2-flag signals (one below the other) exist?

Solution:

Step 1 — (i) Chain the tosses. Each toss has 2 outcomes and repetition is inherent: $2 \times 2 \times 2 = 2^3 = 8$ sequences.

Step 2 — (ii) Fill the two flag places. Upper flag 5 ways, lower flag 4 ways: $5 \times 4 = 20$ signals.

Takeaway: Independent repeatable stages give powers; distinct-object stages give falling products.