Thread: Combinations

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.

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.

Originally Posted by Stuck Man

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.

The correct answer is some what more complicated than the first reply would suggest.
The letters in are only seven difference letters.
The double complicates matters.
So count all five letter combinations that contain at most one . Then add to it the combinations that contain two l’s.

I can't see why you think the order of the letters in syllabus is important.

Originally Posted by Plato

The correct answer is some what more complicated than the first reply would suggest.
The letters in are only seven difference letters.
The double complicates matters.
So count all five letter combinations that contain at most one . 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.

Originally Posted by Stuck Man

I can't see why you think the order of the letters in syllabus is important.

Nobody claimed that order was important.

Originally Posted by Stuck Man

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.

Your options are....

(a) the selection contains

You need to choose 1 letter from the 4 distinct letters y,a,b,u. The number is

(b) the selection contains , only one

This time choose 2 from the 4 distinct letters.

(c) the selection contains .

(d) the selection contains , only one and only one .

Choose 3 from the 4 distinct letters.

(e) the selection contains , only one and no .

You need to choose all 4 distinct letters.

(f) the selection contains one and no .

(g) the selection contains and no .

(h) the selection contains and no .

The selection must contain an or an .

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.

Here is the easiest way.
The coefficient of in the expansion of is .