Exams

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

 Number of 7 digit numbers = 7!3!×2! = 420

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

=6!3!×2! = 60

Hence the required number of 7 digits numbers = 420 - 60 = 360

Given series is A F Z U G L T _

Starts with A

A + 5 letters = F

go to end of alphabet= Z

Z - 5 letters = U

from F go one more to G + 5 letters = L

from U go one back to U, T - 5 letters = O

Second Method ::

Change the letters to their corresponding number.

A=1

B=2

C=3

...

Z=26

A F Z U G L T _

1 6 26 21 7 12 20 ?

Look at the pattern: the 1st and 3rd numbers = 27, 1+26

the 2nd and 4th numbers = 27, 6+21

the 5th and 7th numbers = 27,

so the 6 and 8th would also equal 27.

27 - 12 = 15,

so the letter is O.