Numbers are to be formed between 1000 and 300 | vedclass

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(B) We need to form $4$-digit numbers between $1000$ and $3000$ using the digits ${1, 2, 3, 4, 5, 6}$ without repetition,such that the number is divis

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$30$

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(B) We need to form $4$-digit numbers between $1000$ and $3000$ using the digits ${1, 2, 3, 4, 5, 6}$ without repetition,such that the number is divisible by $4$.For a number to be divisible by $4$,the number formed by its last two digits must be divisible by $4$.Since the number is between $1000$ and $3000$,the first digit must be $1$ or $2$.Case-$I$: The first digit is $1$.The last two digits can be $12, 16, 24, 32, 36, 52, 56, 64$. Since the first digit is $1$,we exclude pairs containing $1$. The possible pairs are $24, 32, 36, 52, 56, 64$ ($6$ pairs).For each pair,the second digit can be any of the remaining $4 - 2 = 2$ digits (since $4$ digits total,$2$ used in last positions,$1$ used in first position,$1$ left for second position).Wait,let's re-evaluate: The first digit is fixed as $1$. The last two digits are chosen from ${2, 3, 4, 5, 6}$.Possible pairs for last two digits divisible by $4$: $24, 32, 36, 52, 56, 64$ ($6$ pairs).For each pair,the second digit can be any of the remaining $6 - 3 = 3$ digits.So,$6 	imes 3 = 18$ numbers.Case-$II$: The first digit is $2$.The last two digits must be chosen from ${1, 3, 4, 5, 6}$ such that the number is divisible by $4$.Possible pairs: $16, 36, 56, 64$ ($4$ pairs).For each pair,the second digit can be any of the remaining $6 - 3 = 3$ digits.So,$4 	imes 3 = 12$ numbers.Total numbers = $18 + 12 = 30$.

Numbers are to be formed between $1000$ and $3000$,which are divisible by $4$,using the digits $1, 2, 3, 4, 5$ and $6$ without repetition of digits. Then the total number of such numbers is.

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