**Question 64:**

How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

**Explanation:**

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as

Case I: 2, 3, 4, 6

Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways.

Hence total number of ways = 3×6 = 18

Case II: 0, 2, 3, 4

case II a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 × 2 × 2 = 8

Total number of ways in this case = 6 + 8 = 14 ways.

Case III: 0, 2, 4, 6

case III a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 × 2 × 2 = 12

Total number of ways in this case = 6 + 12= 18 ways.

Hence the total number of ways = 18 + 14 + 18 = 50 ways

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