I can't do this question:
Find the number of different selections of 5 letters which can be made from the letters of the word syllabus.
Well, there are 8 letters in the word syllabus. You are being asked to find the number of different selections of 5 letters which can be made from the letters of the word syllabus which is the same as the number of ways one can select 5 letters from the 8 letter set comprised of the 8 letters that form the word "syllabus". So, you can ignore the word "syllabus" and just treat the problem as any other set counting problem.
Since combinations is in your title, I am sure you know how to compute the number of ways to select 5 elements from an 8 element set.
The correct answer is some what more complicated than the first reply would suggest.
The letters in $\displaystyle syllabus$ are only seven difference letters.
The double $\displaystyle ll$ complicates matters.
So count all five letter combinations that contain at most one $\displaystyle l$. Then add to it the combinations that contain two l’s.
There are also two of the letter s; only 6 distinct letters.
I will use uppercase L because lowercase looks like 1.
Case by case like this will give the answer:
no L and 1 s
no L and 2 s's
1 L and no s
1 L and 1 s
1 L and 2 s's
2 L's and no s
2 L's and 1 s
2 L's and 2 s's
Can be made faster realising that 1 L and no s gives the same count as 1 s and no L, etc.
Nobody claimed that order was important.
Your options are....
(a) the selection contains $\displaystyle ssll$
You need to choose 1 letter from the 4 distinct letters y,a,b,u. The number is $\displaystyle \binom{4}{1}=4$
(b) the selection contains $\displaystyle ssl$, only one $\displaystyle l$
This time choose 2 from the 4 distinct letters. $\displaystyle \binom{4}{2}=6$
(c) the selection contains $\displaystyle sll$. $\displaystyle \binom{4}{2}=6$
(d) the selection contains $\displaystyle sl$, only one $\displaystyle s$ and only one $\displaystyle l$.
Choose 3 from the 4 distinct letters. $\displaystyle \binom{4}{3}=4$
(e) the selection contains $\displaystyle s$, only one $\displaystyle s$ and no $\displaystyle l$.
You need to choose all 4 distinct letters. $\displaystyle \binom{4}{4}=1$
(f) the selection contains one $\displaystyle l$ and no $\displaystyle s$. $\displaystyle \binom{4}{4}=1$
(g) the selection contains $\displaystyle ss$ and no $\displaystyle l$. $\displaystyle \binom{4}{3}=4$
(h) the selection contains $\displaystyle ll$ and no $\displaystyle s$. $\displaystyle \binom{4}{3}=4$
The selection must contain an $\displaystyle s$ or an $\displaystyle l$.
Sum all of these.
EDIT: Hi "undefined", we were on the same wavelength!
Now it's broken down, faster solutions can be given and maybe some neat closed-form options..
A slightly simpler way:
Start by only considering the case that there is only 1 L and 1 S. Hence the letters to slelect from are A,B,L,S,U and Y. How many ways can you pick five of these at a time? Answer = C(6,5) = 6. This is the number of ways you can select letters with either 0 or 1 L and with either 0 or 1 S.
Now consider combinations that have 2 L's. How many ways can you select three from the remaining 5 letters A,B,S,U, and Y?: C(5,3)=10. This is the number of selections that have 2 L's and either 0 or 1 S.
Ditto for combinations that have 2 S's: C(5,3) = 10 is the number of selections that have 2 S's and either 0 or 1 L.
Finally add the number of ways you can have 2 L's and 2 S's: you have to select only one letter left from the 4 remaining: C(4,1)=4.
Add them up and you get 30.