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find 4 different digit no. from the digit 0,1,2,3,4,5,6 , which is divisable by 4 and second digit (10th digit) is odd
A number is divisible by 4 if its last two digits are divisible by 4. In our case this means our last two digits can be 12, 16, 32, 36, 52, 56.
If our last two digits are 12 then the first digit can be 3, 4, 5, 6, and the second digit can be 0, 3, 4, 5, 6. This gives us a total of 4*4 = 16 legal numbers that end in 12. This process is the same for the other possible combinations, giving us a final answer of 16*6 = 96.
A second method is to start by counting the permutations of four digits out of the seven possible, which is 840. 1/7 of those begin with 0 and are illegal, so after removing those we have 720 legal four digit numbers. From here we break these numbers down into three different types: those that begin with an odd number, those that begin with 4, and those that begin with 2 or 6. You can see that half of the numbers are of the first type, 1/6 are of the second type, and 1/3 are of the third type.
For the first type of number, since the first digit is odd we know that 1/3 of the numbers have an odd number as the 10's digit. Out of these remaining numbers 2/5 have 2 or 6 as the 1's digit. Overall this gives us 720 * 1/2 * 1/3 * 2/5 = 48 legal numbers that begin with an odd digit.
For numbers that begin with 4, 1/2 have an odd digit in the 10's position, and 2/5 of the remaining numbers end in 2 or 6, giving us a total of 720 * 1/6 * 1/2 * 2/5 = 24 numbers that begin with 4.
For numbers that begin with 2 or 6, 1/2 have an odd digit in the 10's position and 1/5 of the remaining numbers end in 2 or 6, giving us 720 * 1/3 * 1/2 * 1/5 = 24 numbers the begin with 2 or 6.
This gives us a final answer of 48 + 24 + 24 = 96.
Wrap it in bacon
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thank dude
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