All possible 5-digit numbers,each having 5 di | vedclass

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(C) The given set of digits is $S = \{1, 2, 3, 5, 6, 8\}$. The total number of digits is $6$. We need to form a $5$-digit number using $5$ distinct di

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) The given set of digits is $S = \{1, 2, 3, 5, 6, 8\}$. The total number of digits is $6$. We need to form a $5$-digit number using $5$ distinct digits from $S$. First,we find the sum of all $6$ digits: $1 + 2 + 3 + 5 + 6 + 8 = 25$. To form a $5$-digit number,we must exclude one digit. Let the excluded digit be $x$. The sum of the remaining $5$ digits will be $25 - x$. For the number to be divisible by $3$,the sum of its digits must be divisible by $3$. If $x = 1$,sum $= 24$ (divisible by $3$). If $x = 4$ (not in set),$x = 2$,sum $= 23$ (not divisible). If $x = 3$,sum $= 22$ (not divisible). If $x = 5$,sum $= 20$ (not divisible). If $x = 6$,sum $= 19$ (not divisible). If $x = 8$,sum $= 17$ (not divisible). Thus,the only set of $5$ digits whose sum is divisible by $3$ is $\{2, 3, 5, 6, 8\}$. The number of $5$-digit numbers formed using these $5$ digits is $5! = 120$. For a number to be divisible by $6$,it must be even and divisible by $3$. Since all these numbers are divisible by $3$,we only need to count those that are odd (to ensure they are not divisible by $6$). The odd digits in the set $\{2, 3, 5, 6, 8\}$ are $\{3, 5\}$. If the last digit is $3$,the remaining $4$ positions can be filled in $4! = 24$ ways. If the last digit is $5$,the remaining $4$ positions can be filled in $4! = 24$ ways. Total numbers divisible by $3$ but not by $6$ $= 24 + 24 = 48$.

All possible $5$-digit numbers,each having $5$ distinct digits,are formed using the digits $\{1, 2, 3, 5, 6, 8\}$. Among them,the number of such numbers which are divisible by $3$ but not by $6$ is:

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