Can someone answer. there is only one problem. there is a picture. !

The number of permutations of the 25 letters taken 2 at a time (with repetitions) is:
The number of permutations of the 9 digits taken 4 at a time (with repetitions) is:

Each permutation of letters can be taken with each permutation of digits, therefore the total number of possible passwords is:

If we have " n "  items and we want to arrange " r " items, the permutation if repetition is allowed =    
                                       

We have 10 digits, since we are not using 0, then we have 9, n= 9.

Since we want to make 4 digits, so permutation with repetition allowed

=  9⁴ = 6561

We have 26 digits, since we are not using M, then we have 25 left, n= 25.

Since we want to make 2 letters, so permutation with repetition allowed

=  25² = 625

Hence total permutation =  6561 × 625 = 4 100 625

Option C.

But wait a minute!

The question forgot to take note that Alphabets could be capital letters or small letters...

In that case total alphabets of small and big = 2*26 = 52

If we remove the M = 52 - 1 = 51   

In this case = 51² = 2601

Total permutations = 6561 × 2601 = 17 065 161

Well you can challenge the question if you want, but if you just ignore this error from the question and be in the mind of the makers of the question, then option C like as we have showed earlier is the answer.

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.