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Question 30 : 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? [TITA]

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We have to find the no. of four digit numbers which are divisible by 6 that are to be formed using the digits 0, 2, 3, 4, 6.

We should know that 0 does not occur in the left most position. If the numbers are to be divisible by 6 it should be a multiple of 2 and 3.

So it should be an even number and also should be a multiple of 3.

⟹ 2 + 3 + 4 + 6 + 0 = 15

Here we have two choices, drop something such that the number should be a multiple of 3 or we can drop a multiple of 3.

The following are the possibilities,

Drop 3 ⟹ 2 + 4 + 6 + 0 = 12 which is a multiple of 3.

Drop 6 ⟹ 2 + 3 + 4 + 0 = 9 which is a multiple of 3.

Drop 0 ⟹ 2 + 3 + 4 + 6 = 15 which is a multiple of 3.

We have to check the possible outcomes,

For 2 , 4 , 6 , 0 ⟹ __3__ × __3__ × __2__ × __1__ = 18 ways.

For 2 , 3 , 4 , 0 ⟹ __3__ × __3__ × __2__ × __1__ = 18 ways.

But out of these 18 outcomes there will be some possibilities where the last digit is 3.

4 such possibilities exist here which needs to be eliminated.

Therefore 18 – 4 = 14 ways.

For 2 , 3 , 4 and 6 if we have last digit as 3 ie. _ _ _ 3, we can rearrange 2 , 4 and 6 in 3! ways

Hence 2 , 3 , 4 and 6 can be arranged in 4! - 3! = 24 – 6 = 18 ways.

Therefore, in total we can form 18 + 14 + 18 = 50 four digit numbers.

The question is **"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? [TITA]" **

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