The digits 1,2,3,4,5,6,7,8 and 9 are written in random order to form a nine digit number. The probability that this number is divisible by 4, isA. `(1)/(9)`B. `(2)/(3)`C. `(2)/(9)`D. `(7)/(9)`

The digits 1,2,3,4,5,6,7,8 and 9 are written in random order to form a

1.	The digits 1,2,3,4,5,6,7,8 and 9 are written in random order to form a nine digit number. The probability that this number is divisible by 4, isA. `(1)/(9)`B. `(2)/(3)`C. `(2)/(9)`D. `(7)/(9)`
Answer» Correct Answer - C There are `.^(9)P_(9)=9!` ways of arranging the given digits to form a nine digit number. So, total number of 9 digit numbers formed by the given digits =9!. Out of these 9! Numbers only those numbers are divisble by 4 which have their last digits as even natural number and the numbers formed by their last two digits are divisible by 4. The various possibilities of last two digits are : 12, 32, 52, 72, 92 24, 64, 84 16, 36, 56, 76, 96 28, 48, 68 This means that there are 16 ways of choosing the last two digits. Corresponding to each of these ways the remaining 7 digits can be arranged in 7! ways. Therefore, the total number of 9 digit numbers divisble by 4 is `16xx7!` Hence, required probability `=(16xx7!)/(9!)=(2)/(9)`

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The digits 1,2,3,4,5,6,7,8 and 9 are written in random order to form a nine digit number. The probability that this number is divisible by 4, isA. `(1)/(9)`B. `(2)/(3)`C. `(2)/(9)`D. `(7)/(9)`

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