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Question
Assertion :Total number of five-digit numbers having all different digits and divisible by 4 can be formed using the digits {1,3,2,6,8,9} is 192. Reason: A number of divisible by 4, if the last two digits of the number are divisible by 4.
Assertion :Total number of five-digit numbers having all different digits and divisible by 4 can be formed using the digits {1,3,2,6,8,9} is 192. Reason: A number of divisible by 4, if the last two digits of the number are divisible by 4.
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Solution
The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
For the number to be divisible by 4, the last two digits must be a multiple of 4.
Hence
The last two digits can be
Case 1
Last two digits is 12.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 32
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 92.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Hence total =3.4!=72 ways.
Case II
Last two digits is 16.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 36.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 96.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Total =3.4!=72 ways.
Case II
Last two digits is 28.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 68.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Hence total =2.4!=48.
Hence the required permutation is
=72+72+48
=192 ways.
Hence total numbers =192.
For the number to be divisible by 4, the last two digits must be a multiple of 4.
Hence
The last two digits can be
Case 1
Last two digits is 12.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 32
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 92.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Hence total =3.4!=72 ways.
Case II
Last two digits is 16.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 36.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 96.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Total =3.4!=72 ways.
Case II
Last two digits is 28.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Last two digits is 68.
Number of ways of filling the rest 3 places is (6−2)! ways
=4!.
Hence total =2.4!=48.
Hence the required permutation is
=72+72+48
=192 ways.
Hence total numbers =192.
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