# CAT 2017 Question Paper | Quants Slot 2

###### CAT Previous Year Paper | CAT Permutation and Combination Questions | Question 30

Permutation and Combination is one of the topics that students generally fear. But does this topic even need the amount of attention and time that we otherwise allocate to it in our online CAT Preparation?
2IIM has analyzed CAT previous year paper to see the pattern followed and the type of questions you can expect in your upcoming CAT Exam. Visit 2IIM's CAT Blueprint to go through our analysis and plan your CAT preparation accordingly. If you are still not sure where to start from, going through CAT Previous Year Papers can be a good idea. Here is a PnC question from CAT 2017 Slot 2. Solve and see whether you can answer it correctly.

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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##### Method of solving this CAT Question from Permutation and Combination

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]"

##### Hence, the answer is 50

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