Hi there,
Thanks for the post. This helps!
I think there is a problem in Q37.
37. 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
Solution provided is:
The best answer is A.
The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2),
the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6=2520.
I have problems with explanation in
RED.
If on second and third position the number is repeated (3,5, or 7), then fourth position has
8 options and fifth position has
7 options.
So according to me, solution should be 4*2*3*7*6 + 4*3*3*8*7
Please correct me if i am wrong.
Pranav
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