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.

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- Sep 1st 2010, 10:02 AMStuck ManCombinations
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. - Sep 1st 2010, 10:12 AMDanneedshelp
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. - Sep 1st 2010, 10:28 AMStuck Man
8C5 is 56. It is possible to select S twice or L twice or two S's and two L's. I think I need to subtract something from 56. The answer in the book is 30.

- Sep 1st 2010, 10:43 AMPlato
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. - Sep 1st 2010, 10:57 AMStuck Man
I can't see why you think the order of the letters in syllabus is important.

- Sep 1st 2010, 11:39 AMundefined
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. - Sep 1st 2010, 11:43 AMArchie Meade
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.. - Sep 1st 2010, 01:17 PMebaines
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. - Sep 1st 2010, 01:31 PMPlato
Here is the easiest way.

The coefficient of $\displaystyle x^5$ in the expansion of $\displaystyle \displaystyle \left( {\sum\limits_{k = 0}^1 {x^k } } \right)^4 \left( {\sum\limits_{k = 0}^2 {x^k } } \right)^2 $ is $\displaystyle 30$.