The total number of numbers lying between 100 | vedclass

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(A) We need to form $3$-digit numbers using the set $\{1, 2, 3, 4, 5\}$ without repetition.$1$. Divisibility by $3$: $A$ number is divisible by $3$ if

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$32$

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(A) We need to form $3$-digit numbers using the set $\{1, 2, 3, 4, 5\}$ without repetition.$1$. Divisibility by $3$: $A$ number is divisible by $3$ if the sum of its digits is divisible by $3$. The possible triplets from $\{1, 2, 3, 4, 5\}$ whose sum is a multiple of $3$ are:$(1, 2, 3) ightarrow 3! = 6$ numbers$(1, 3, 5) ightarrow 3! = 6$ numbers$(2, 3, 4) ightarrow 3! = 6$ numbers$(3, 4, 5) ightarrow 3! = 6$ numbersTotal divisible by $3 = 6 + 6 + 6 + 6 = 24$.$2$. Divisibility by $5$: $A$ number is divisible by $5$ if its last digit is $5$. The possible $3$-digit numbers are of the form $XY5$,where $X, Y \in \{1, 2, 3, 4\}$.The number of such arrangements is $4 	imes 3 = 12$.$3$. Divisibility by both $3$ and $5$: These are numbers ending in $5$ whose sum of digits is a multiple of $3$. The possible sets are $\{1, 3, 5\}$ and $\{3, 4, 5\}$.For $\{1, 3, 5\}$,numbers ending in $5$ are $135$ and $315$ ($2$ numbers).For $\{3, 4, 5\}$,numbers ending in $5$ are $345$ and $435$ ($2$ numbers).Total divisible by both $= 2 + 2 = 4$.$4$. Using the Principle of Inclusion-Exclusion:Total $= n(3) + n(5) - n(3 \cap 5) = 24 + 12 - 4 = 32$.

The total number of numbers lying between $100$ and $1000$ that can be formed with the digits $1, 2, 3, 4, 5$,if the repetition of digits is not allowed and the numbers are divisible by either $3$ or $5$,is:

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