Problem Solving - 19

How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500


Answer: Given OA - A is incorrect ... ans is 2688 which is not an option in the choices...

When 2nd and 3rd digit gets repeated:

The first digit will be a non zero even (2, 4, 6, 8) = 4 ways
3rd digit is a non even prime = (3, 5, 7) = 3 ways
2nd digit is a REPEAT of that prime: 1 way
the fourth digit has not been used: 8 ways
the fifth digit has not been used: 7 ways

Hence 4*3*8*7 = 672 ways

Now, the non repeating case:

1st digit will be a non zero even (2, 4, 6, 8) = 4 ways
3rd digit (3, 5, 7) = 3 ways
2nd (no repeat and odd) = 4 ways
4th digit = 7
5th digit = 6

Hence 4*3*4*7*6 = 2016

Total number of ways = 2016 + 672 = 2688 ways


Note:
1). 1 is neither a prime nor a composite number.
2). 2 is the only even prime number.