(A) The available odd digits are $\{1, 3, 5, 7, 9\}$. There are $5$ such digits.To form a $6$-digit number using only these $5$ digits such that all d

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(A) The available odd digits are $\{1, 3, 5, 7, 9\}$. There are $5$ such digits.To form a $6$-digit number using only these $5$ digits such that all digits appear at least once,exactly one digit must be repeated.Step $1$: Select the digit to be repeated from the $5$ available odd digits in $^5C_1 = 5$ ways.Step $2$: Arrange these $6$ digits (where one digit is repeated twice) in $\frac{6!}{2!}$ ways.Total numbers = $5 \times \frac{6!}{2!} = 5 \times \frac{720}{2} = 5 \times 360 = 1800$.

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

Medium

How many $6$-digit numbers can be formed using only odd digits,such that all odd digits appear at least once?

A
$5 \times \frac{6!}{2!}$
B
$6!$
C
$\frac{1}{2} \times 6!$
D
None of these

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

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How many $6$-digit numbers can be formed using only odd digits,such that all odd digits appear at least once?

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