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How many distinct 'numbers' (with REPETITION ALLOWED) can be formed using the digits 0, 1, 2, 3 in 4 places (the result can be a string like 0023)?
A$P(4, 4) = 24$
B$\binom{4}{4} = 1$
C$4!$
D$4^4 = 256$
Answer & Solution
Correct answer: D. $4^4 = 256$
1. With REPETITION ALLOWED, each of the 4 positions independently has 4 choices (digits 0, 1, 2, 3).
2. By multiplication principle: $4 \cdot 4 \cdot 4 \cdot 4 = 4^4 = 256$.
3. Examples: 0000, 0001, 1234, 3333, ... — all are valid strings.
4. Option A treats them as distinct without repetition. Option B is wildly wrong. Option D is the permutation of 4 distinct elements.
5. Key principle: when each position is INDEPENDENT and repetition is allowed, use $n^r$ (where $n$ is the number of choices per position, $r$ is the number of positions).
_Source: NCERT Class 11 Mathematics, Ch 6, §6.3.2 (Permutations with repetition), p. 7._
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