Answer with explanation:
→→(A):
Number of digits required to make or open a lock = 3→ [1 letter +2 number]
Number of letters from which we have to choose , 1 Letter = 4={A,B,C,D}
Number of digits from which we have to choose a digit =2={5,6}
Size of the Sample Space
=Choosing a letter from 4 letters × Placing 2 numbers that is 5 and 6.
={(A,5,6),(B,5,6),(C,5,6),(D,5,6)}
→Each of the 3 digit lock has ,3! or

=6, distinct arrangement.
So, total Size of sample Space=6 × 4=24 ⇒→Option D
→→(B):
There is one odd number, that is, 5,so the odd number must come in between the three digit code, Out of 24, sample space .
So, Favorable Outcome={(6 5 A),(6 5 B),(6 5 C),(6 5 D)}=4
Required Probability

Option B

→→(C):
Out of 24 , three digit codes, number of codes in which D is in the code but is not in the first position ={(5,D,6),(6,D,5),(5,6,D),(6,5,D)}=4
Required Probability

Option B