(B) number is divisible by $10$ if its units digit is $0$.Since the number must be a $6$-digit number and we have $6$ distinct digits $(0, 1, 3, 5, 7,

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(B) number is divisible by $10$ if its units digit is $0$.Since the number must be a $6$-digit number and we have $6$ distinct digits $(0, 1, 3, 5, 7, 9)$,the digit $0$ must be fixed at the units place.This leaves $5$ remaining positions to be filled by the remaining $5$ digits $(1, 3, 5, 7, 9)$.The number of ways to arrange these $5$ digits in the $5$ vacant places is given by $5!$.$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.Thus,the total number of such $6$-digit numbers is $120$.

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 from the digits $0, 1, 3, 5, 7,$ and $9$ which are divisible by $10$,given that no digit is repeated?

A
$96$
B
$120$
C
$144$
D
$720$

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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 $6$-digit numbers can be formed from the digits $0, 1, 3, 5, 7,$ and $9$ which are divisible by $10$,given that no digit is repeated?

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