Thank you. Can you please help me with 16b too?

This method works as all the letters of BYTES are unique.

If it were BBTES ,BEETY or something like BYTYB you will need to eliminate the repeated cases.

16b


3 distinct letters, 2 distinct digits, the letters must come first.

To visualise, we can use an example like
ABC12


So we already know the positions of the letters can only be the first 3 characters and the digits can only be at the last 2 characters.


Number of ways to permutate the 3 distinct letters
= 3! = 6

Or

① 3 ways for the 1st letter (A,B,or C)

② For each of the 3 ways, there are 2 ways for the 2nd letter since there are only 2 letters remaining.

If A was first letter, 2nd letter can be B or C
If B was first letter, 2nd letter can be A or C
If C was first letter, 2nd letter can be A or B


③ Only 1 letter is left so 1 way.

Number of ways = 3 x 2 x 1 = 6
(This is basically 3!)



Similarly,


Number of ways to permutate the 2 digits
= 2! = 2

Or

① 2 ways for the 1st digit (1 or 2)
② For either of the 2 ways, there is only 1 digit left. So 1 way for the 2nd digit

Number of ways = 2 x 1 = 2
(This is basically 2!)


Total number of ways
= number of ways for 3 letters x number of ways for 2 digits
= 6 x 2
= 12

To confirm (for understanding) the full list will be :

ABC12
ACB12
BAC12
BCA12
CAB12
CBA12

ABC21
ACB21
BAC21
BCA21
CAB21
CBA21