(C) We need to form $4$-digit numbers using the digits ${0, 2, 3, 6, 7, 8}$.Total permutations of $4$ digits chosen from $6$ is $^6P_4 = \frac{6!}{2!}

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(C) We need to form $4$-digit numbers using the digits ${0, 2, 3, 6, 7, 8}$.Total permutations of $4$ digits chosen from $6$ is $^6P_4 = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360$.However,a $4$-digit number cannot have $0$ in the thousands place.The number of permutations where $0$ is in the thousands place is $^5P_3 = \frac{5!}{2!} = 5 \times 4 \times 3 = 60$.Therefore,the total number of valid $4$-digit numbers is $360 - 60 = 300$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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How many numbers between $999$ and $10,000$ can be formed using the digits $0, 2, 3, 6, 7, 8$ without repetition?

A
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B
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C
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D
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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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How many numbers between $999$ and $10,000$ can be formed using the digits $0, 2, 3, 6, 7, 8$ without repetition?

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