Four children are running a race. (i) In how many ways can the first two places be filled? (ii) In how many di

Exercise 4.1 · Q3

Q. Four children are running a race.

(i) In how many ways can the first two places be filled?

(ii) In how many different ways could they finish the race?

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Concept understanding — Fundamental Principles of Counting

The Fundamental Principle of Counting

Imagine you're getting dressed. You have 3 shirts (red, blue, green) and 2 pairs of pants (black, white). How many different outfits can you make?

You could list them: red-black, red-white, blue-black, blue-white, green-black, green-white. That's 6 outfits.

Notice something: 3 shirts × 2 pants = 6 outfits. That's not a coincidence — it's the entire idea.

The Intuition: A Fork in the Road

Think of each choice as a fork in a path. First you pick a shirt: 3 options. For each shirt, you then pick pants: 2 options. So the total number of paths through the whole process is 3 × 2 = 6.

This works no matter how many stages you add. If you also had 2 pairs of shoes, the total outfits become 3 × 2 × 2 = 12. Each new choice multiplies the total.

Note

The key insight: you multiply, not add. Addition would mean you choose either a shirt or pants, not both. Multiplication means you choose one of each, in sequence.

The Precise Statement

Fundamental Principle of Counting (Multiplication Principle)

If an event can happen in $m$ ways, and after it happens a second event can happen in $n$ ways, then the two events together can happen in $m \times n$ ways.

More generally: if a task consists of $k$ steps, and step 1 can be done in $n_1$ ways, step 2 in $n_2$ ways, ..., step $k$ in $n_k$ ways, then the total number of ways to complete the entire task is:

n_1 \times n_2 \times \cdots \times n_k

Watch out

This principle only works when the number of ways for each step does not depend on which choice was made earlier. If picking a red shirt somehow limits your pants options, you cannot simply multiply — you'd need to count more carefully.

A Simple Example

How many 2-digit numbers can you form using the digits 1, 2, 3, 4, 5 if repetition is allowed?

Step 1: Choose the tens digit — 5 options (1,2,3,4,5)
Step 2: Choose the units digit — 5 options (1,2,3,4,5)

Total = $5 \times 5 = 25$ numbers.

If repetition were not allowed:

Step 1: 5 options
Step 2: only 4 options (since one digit is already used)

Total = $5 \times 4 = 20$ numbers.

Tip

When repetition is not allowed, the number of options decreases by 1 at each step. When repetition is allowed, the number stays the same.

Why This Matters

This principle is the foundation of all counting problems in combinatorics. Permutations, combinations, arrangements — every one of them is just a special case of this multiplication idea, with some extra rules about repetition and order.

The rule is simple: when choices happen in sequence, multiply the number of options at each step.

Both ask for an ordered selection from the same 4 children.

Step 1. (i) First place: $4$ choices; second place: $3$ remaining choices $\Rightarrow {}^4P_2=4\times3=12$ .

Step 2. (ii) All 4 children finish in some order (a full ranking): $4!=24$ .

✓Final answer

12

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of 5 digit numbers, all digits of which are odd is:$
(a) $5^6$
(b) $25$
(c) $625$
(d) $5^5$
›Reveal solutionSolution

Each of the 5 digit-positions is filled independently from the 5 odd digits.

The odd digits available are $\{1,3,5,7,9\}$ , a set of 5 digits. Since none of them is $0$ , there is no restriction on the leading digit either. Each of the 5 positions of the number can be filled in 5 ways, independently of the others, so the total count is

$5\times5\times5\times5\times5=5^5.$

✓Final answer
$5^5$
CBSE 2022Set ANNUAL1 markMCQ
$Q. The number of 5 digit numbers all digits of which are odd is:$
(a) $5^6$
(b) 25
(c) 625
(d) $5^5$
›Reveal solutionSolution

With 5 choices of odd digit for each of the 5 positions (no restriction, since all odd digits are already non-zero), the count is $5^5$ .

The odd digits are $1,3,5,7,9$ -- exactly 5 of them.

A 5-digit number has 5 positions, and unlike a general 5-digit number problem, there's no separate restriction on the leading digit here, because none of the allowed digits (odd digits) is zero anyway.

By the fundamental principle of counting, each of the 5 positions can independently be filled in 5 ways, giving $5\times5\times5\times5\times5 = 5^5$ numbers.

✓Final answer
$5^5$
CBSE 2019Set ANNUAL1 markMCQ
$Q. The number of five digit numbers in which all digits are even, is:$
(a) $4\times 5^4$
(b) $4\times 5^5$
(c) $5^5$
(d) $5\times 5$
›Reveal solutionSolution

With digits restricted to $\{0,2,4,6,8\}$ , the first digit excludes 0 (4 choices) and the remaining 4 digits are unrestricted (5 choices each), giving $4\times 5^4$ .

The even digits available are $0, 2, 4, 6, 8$ — 5 digits in total, and repetition is allowed (the question only restricts which digits may be used, not that they be distinct).

For a 5-digit number, the first (leftmost) digit cannot be 0, so it has $5-1=4$ choices.

Each of the remaining 4 digit-positions can be any of the 5 even digits (0 is allowed here), giving 5 choices each.

By the multiplication principle, the total count is $4\times 5\times 5\times 5\times 5 = 4\times 5^4$ .

✓Final answer
$4\times 5^4$

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Q.Four children are running a race.

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