Math Is Fun Forum

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Counting = Arrangement with restriction

In how many ways can the letters of the word ALTERNATE be arranged so that there is a vowel at each end?
Answer = 7560

I know how to represent this with boxes but only when each object is different

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Counting = Arrangement with restriction

I've founded a quick way to calculate this!
We have 2 A-s, 2 E-s and 2 T-s.
Let us index the same letters, e.a. A1,A2,E1,E2,T1,T2.
Now we'll use the system{A1,A2,E1,E2,T1,T2,L,R,N}.
In this system the words
A1 LTERN A2 TE
and
A2 LTERN A1 TE
are different, but in the original system, they are the same word.
Now, let's meashure the following:
1 word form the normal system = ? words form the extended system?
Any "normal" word won't change, if we substitute A1 with A2 and E1 with E2 and T1 with T2. 3 couples in total , so:
1 word form the normal system =2^3=8 words form the extended system.
Now, we just need to find the word form the extended system and divide by 8.
In the extended systen, we have no couples, at the first place should stand 1 of 4 letters - A1, A2, E1 or E2. at the last place we should place one of the left 3. Between them, we'll have 7 "empty slots" for filling, which could be done in 7! ways. So the answer is:

words in total.

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