SOLUTION: find the number of different selections of 4 letters from the 9 letters of the word HAPPINESS which contains no Ps and one or two Ss.

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Question 1075402: find the number of different selections of 4 letters from the 9 letters of the word HAPPINESS which contains no Ps and one or two Ss.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
The word "selection" is ambiguous. Does order matter or not?

For instance,

Is SANE the same 'selection' as EANS, or are they the same?
Is HISS the same 'selection' as SISH, or are they the same?

I will assume that in those two cases they are different.

So I'll assume you meant "arrangements" or "permutations", where 
order matters.  If you meant that order does not matter, then 
if you'll tell me in the thank-you note form below, I'll get 
back to you by email.  (No charge ever)

Since there are no P's, we are selecting from HAINESS

Case 1:  Those that contain exactly 1 S

We are selecting 4 letters from HAINES

Choose the 1st letter any of 6 ways.
Choose the 2nd letter any of the remaining 5 ways.
Choose the 3rd letter any of the remaining 4 ways.
Choose the 4th letter any of the remaining 3 ways.

That's 6×5×4×3 = 360

Case 2: Those that contain exactly 2 S's

Choose the 2 non-S's from HAINE.  
That's "5 choose 2" or 5C2 = 10 ways.

For each of those 10 choices we have the same number
of permutations as the number of distinguishable 
permutations of XYSS, where the X and Y stand for the 
2 non-S's we chose.  That would be 4! if the S's were 
distinguishable.  But since the 2 S's are not 
distinguishable,  we must divide by 2!. That's 
4!/2! = 24/2 = 12 ways.

That's 10×12 = 120 for case 2

Total for both cases:  360+120 = 480

Edwin