codes and outcomes finite

Question

Assume that product codes are formed from the letters S, Y, R, and V, and consist of 5 not necessarily distinct letters arranged one after the other. For example, SSSYS is a product code.

 

How many different product codes contain exactly one S

Answer 1

S, Y, R, and V are used to make 5-letter codes

you want only one S

there are 5 positions

S can be in position 1, for position 2 you can choose from 3 letters, for position 3 you can choose from 3 letters, for position 4 you can choose from 3 letters, and for position 5 you can choose from 3 letters

there are 3*3*3*3=81 codes with S in position 1

S can be in position 2, for positions 1, 3, 4, 5 you can choose from 3 letters for each position

3*3*3*3=81 codes with S in position 2

S can be in position 3

there are 81 codes with S in position 3

there are 81 codes with S in position 4

there are 81 codes with S in position 5

there are 5*81=405 codes

S3333

3S333

33S33

333S3

3333S

 

 

Answer 2

Hopefully, other tutors will give an equation and explain the math (combinations) for this problem.

However, I’ll give some analysis:  If there is to be one and only one S (“exactly one S”), we could take the liberty of sorting the product codes just for purposes of counting them.  So, if the first letter must be an S and there may not be another S in the code, we simply have 4 more letters in the code.

How many combinations are there of Y, R, and V (no S) in 4 positions?  It’s like a gas pump (or odometer, or bicycle lock, etc.) with 4 dials that can have any of 3 values:  3*3*3*3 (=81) possibilities.
 
There are also 81 possibilities when S is in position 2; another 81 possibilities when S is in position 3; and 81 possibilities when S is 4, and another 81 possibilities when S is in position 5.  Although not in alphabetic code order, this accounts for all codes (and counts those with exactly one S).  Thus, 81+81+81+81+81 = 405 total.