Once you attempt the question then PrepInsta explanation will be displayed.

First we fix 1 at unit place of a six digit number.

Since any number can be divided by 1 .

So it does not matter what comes at other five places .

_ _ _ _ _ 1

So, first position can be filled in 5 ways as any number

from 2 to 6 can be placed in that position .

Similarly 2nd position can be filled with 4 ways .

So , by fundamental principle of counting ,

We get,

5x4x3x2x1x1=120 or 5! x 1=120

Now for 2 at units place ,

Any even number is divided by 2 ,

Hence it is again similar to first case,

5! × 1 =120

Now for 3 at units place ,

For any number to be divisible by 3 , we need to make

sure that sum of the digits is divisible by 3.

Now , 1+ 2 + 3 +4 + 5 +6 = 21

Since 21 is divisible by 3 , so any number formed by number

1 to 6 is divisible by 3 .

Hence , 5! × 1 =120.

Similarly for 5 , 5! ×1 = 120

Now for 6 at unit place ,

We know that any number which is divisible by 2 and

3 is also divisible by 6 .

Hence , 5! ×1 =120

Now for 4 at unit place.

If there is even number at tens place with 4 at units place ,

then the number is divided by 4 else no.

So we have only two even number 2 and 6 which we can

put at tens place .

So now first fixing 2 at tens place and 4 at units place .

_ _ _ _ 2 4

We get by fundamental principle of counting ,

4 × 3 × 2 × 1 × 1 ×1

Or

4! × 1 ×1 = 24

Similarly for 6 at tens place and 4 at units place .

We get 4! × 1 × 1 = 24

Now let us add all the values we got ,

120 +120 +120 + 120 +120 +24 +24 = 648

Hence answer is (c)

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