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Question

Consider the set S={0,1,2,3,,9}. Let m denotes the number of ways the two numbers a,b with replacement chosen from S such that |ab| >6 and n denotes the number of 10 digit numbers that can be formed using each and every digit of S, which are divisible by 2 but not by 3, then which of the following is/are correct?

A
m+n=12
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B
m+n=16
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C
mn=12
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D
mn=8
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Solution

The correct options are
A m+n=12
C mn=12
For m, |ab|>6
|ab|7

ab07,8,93 ways 18,9 2 ways 29 1 way 70 1 way 80,1 2 ways 90,1,23 ways
Number of ways m=2(1+2+3)=12 ways

S={0,1,2,3,,9}
Sum of the digits is 9×102=45 which is divisible by 3.
So, every 10 digit number is always divisible by 3.
n=0

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