Question

Can someone please answer. There is only one problem. There is a picture. Thank you!!!

Answer 1

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

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.

Answer 2

The number of permutations of the 25 letters taken 2 at a time (with repetitions) is:$25^{2}$
The number of permutations of the 9 digits taken 4 at a time (with repetitions) is:
$9^{4}$
Each permutation of letters can be taken with each permutation of digits, therefore the total number of possible passwords is:
$25^{2}\times 9^{4}=4,100,625$