**Quantitative Aptitude – Modern Maths – Permutation and Combination**

**Question**

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?

**Answer**

50

**Solution**

From CAT 2017 – Quantitative Aptitude – Modern Maths – Permutation and Combination, we can see that,

For a number to be divisible by 6, it should have both 2 and 3 as factors.

For 3: The sum of digits should be divisible by 3

For 2: The last digit of the number should be even

There can be three such cases, (0,2,3,4), (0,2,4,6) and (2,3,4,6)

1st case: (0,2,3,4)

When the last digit is ‘0’

The number of combinations can be 3! = 6

When the last digit is 2/4

The number of combinations will be 2*2*1*2 = 8

Total = 14

2nd case: (0,2,4,6)

When the last digit is ‘0’

The number of combinations = 3! = 6

When the last digit is 2/4/6

The number of combinations = 2*2*1*3 = 12

Total = 18

3rd case: (2,3,4,6)

Last digit has to be 2/4/6

So, the number of combinations = 3*2*1*3 = 18

Total number of combinations = 14+18+18 = 50

Answer: 50

## Download CAT 2017 Question Paper with answers and detailed solutions in PDF

## CAT 2017 Questions from Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – P&C – Q1: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

Quantitative Aptitude – Modern Maths – P&C – Q3: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is

Quantitative Aptitude – Modern Maths – Progressions – Q2: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.

Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?

Quantitative Aptitude – Modern Maths – Progressions – Q4: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

Quantitative Aptitude – Modern Maths – Progressions – Q5: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is

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