If all the possible 3-digit numbers are forme | vedclass

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(D) number is divisible by $3$ if the sum of its digits is divisible by $3$. The given digits are $S = \{1, 3, 5, 7, 9\}$.We need to choose $3$ digits

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) number is divisible by $3$ if the sum of its digits is divisible by $3$. The given digits are $S = \{1, 3, 5, 7, 9\}$.We need to choose $3$ digits from $S$ such that their sum is a multiple of $3$.The possible sets of $3$ digits are:$1) \{1, 3, 5\} ightarrow 	ext{Sum} = 9$ (Divisible by $3$)$2) \{1, 5, 9\} ightarrow 	ext{Sum} = 15$ (Divisible by $3$)$3) \{3, 5, 7\} ightarrow 	ext{Sum} = 15$ (Divisible by $3$)$4) \{5, 7, 9\} ightarrow 	ext{Sum} = 21$ (Divisible by $3$)$5) \{1, 3, 9\} ightarrow 	ext{Sum} = 13$ (Not divisible by $3$)$6) \{1, 7, 9\} ightarrow 	ext{Sum} = 17$ (Not divisible by $3$)$7) \{3, 7, 9\} ightarrow 	ext{Sum} = 19$ (Not divisible by $3$)$8) \{1, 3, 7\} ightarrow 	ext{Sum} = 11$ (Not divisible by $3$)$9) \{1, 5, 7\} ightarrow 	ext{Sum} = 13$ (Not divisible by $3$)$10) \{3, 5, 9\} ightarrow 	ext{Sum} = 17$ (Not divisible by $3$)There are $4$ such sets of $3$ digits whose sum is divisible by $3$.Each set of $3$ distinct digits can be arranged in $3! = 6$ ways.Total numbers $= 4 	imes 6 = 24$.

If all the possible $3$-digit numbers are formed using the digits $1, 3, 5, 7, 9$ without repeating any digit,then the number of such $3$-digit numbers which are divisible by $3$ is

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