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Possible numbers in each column

1: 2,4,6,8 (cannot be 0 because has to be 4 digit number)
2. 1,3,5,7,9
3. 2,3,5,7
4. 0,3,6,9 (it's easy to forget the 0 here)

It is now easiest to imagine 3 scenarios:

1. a 2 appears in the first column and no other column.
That means there is only 1 way of arranging the first column, 5 ways the second, 3 ways the third, and 4 ways the fourth.
1x5x3x4 =60

2. a 2 appears in the third column and no other column
That means there are 3 ways of arranging the first column, 5 ways the second, 1 way the third, and 4 ways the fourth.
3x5x1x4 =60

3. no 2s appear at all
That means there are 3 ways of arranging the first column, 5 ways the second, 3 ways the third, and 4 ways the fourth.
3x5x3x4 =180

Next, add all the scenarios:
60+60+180 =300

D
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As per me the answer should be 225 as in the fist digit 0 cannot come .

So i have done following:-

1st Digit--> 2,4,6,8
2nd Digit-->1,3,5,7,9
3rd Digit-->2,3,5,7
4th Digit--> 3,6,9

So if i fix 2 in 1000th positon i have 5X3X3=45 possibility.
now if i Fix 4 i have--> 5x4x3=60
Same for 6&8-60

total-45+60+60+60=225
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SahilGera18
As per me the answer should be 225 as in the fist digit 0 cannot come .

So i have done following:-

1st Digit--> 2,4,6,8
2nd Digit-->1,3,5,7,9
3rd Digit-->2,3,5,7
4th Digit--> 3,6,9

So if i fix 2 in 1000th positon i have 5X3X3=45 possibility.
now if i Fix 4 i have--> 5x4x3=60
Same for 6&8-60

total-45+60+60+60=225

I think you forgot to include the 0 in the units column, because there are always 4 numbers there that are divisible by 3: 0,3,6,9

Do you mind sharing why you fixed the 4, 6, and 8?
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I fixed 4,6,8 because the thousand have to even so total 2,4,6,8

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Bunuel
How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

A. 20
B. 150
C. 225
D. 300
E. 320

Possibilities for the first digit = 2, 4, 6, 8
Possibilities for the second digit = 1, 3, 5, 7, 9
Possibilities for the third digit = 2, 3, 5, 7
Possibilities for the fourth digit = 0, 3, 6, 9

Total possibilities = 4 * 5 * 4 * 4 = 320
Number of cases where both the 1st digit and the 3rd digit is 2 = 1 * 5 * 1 * 4 = 20

Thus, number of cases where 2 is used only once = 320 - 20 = 300
Answer D
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