How many three digit numbers can be formed using the digits 0, 1, 3, 5 and 8. (i) If repetition of digits is n

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Q. How many three digit numbers can be formed using the digits 0, 1, 3, 5 and 8.

(i) If repetition of digits is not allowed

(ii) If repetition of digits is allowed.

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Fill the hundreds, tens and units places by the multiplication principle, remembering the hundreds digit cannot be $0$ .

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Fundamental principle of counting: if a task is done in stages with $m_1, m_2, m_3$ choices respectively, the total number of ways $= m_1 \times m_2 \times m_3$ . The leftmost (hundreds) digit of a number can never be $0$ .

Available digits: $\{0,1,3,5,8\}$ , so $n=5$ digits in all.
(i) Repetition not allowed:
- Hundreds place: cannot be $0$ $\Rightarrow$ $4$ choices ( $1,3,5,8$ ).
- Tens place: any digit except the one already used $\Rightarrow$ $4$ choices (now $0$ is allowed).
- Units place: any digit except the two already used $\Rightarrow$ $3$ choices.
- Total $= 4 \times 4 \times 3 = 48$ .
(ii) Repetition allowed: …

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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, independ …

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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.

…

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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.

…

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Q.How many three digit numbers can be formed using the digits 0, 1, 3, 5 and 8.

Concept understanding — Fundamental Principles of Counting

The Fundamental Principle of Counting

The Intuition: A Fork in the Road

The Precise Statement

A Simple Example

Q.How many three digit numbers can be formed using the digits 0, 1, 3, 5 and 8.

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