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-   -   Further Premutation / Combination problem (http://www.physicsforums.com/showthread.php?t=581239)

kenny1999 Feb25-12 01:02 PM

Further Premutation / Combination problem
 
1. The problem statement, all variables and given/known data

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

(This is a question with solution I find on web, but I don't understand the solution)

2. Relevant equations



3. The attempt at a solution

The following is the modal answer to the problem, but I don't understand why
7! / 2! and 5! / 3!. It's very hard to think!






Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters = 7! / 2! = 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in 5! / 3! = 20 ways.

Required number of ways = (2520 x 20) = 50400.

drawar Feb26-12 01:34 AM

Re: Further Premutation / Combination problem
 
This is called permutation with repetition.
Take 'APPLE' for example, the word contains 2 identical P's but we can distinguish them by adding suffixes, i.e. A(P1)(P2)LE.
Then the number of permutation of the letters A, P1, P2, L, E is 5!
But this number includes separately the 2 permutations A(P1)(P2)LE and A(P2)(P1)LE, so that the arrangement APPLE is counted twice in the 5! permutations.
Because P1 and P2 can be arranged in 2! ways, every distinct arrangement of the letters A, P1, P2, L, E is included 2! in the 5! permutations.
Therefore the numbers of arrangements of the letters A, P, P, L, E is 5!/2!
Things should be clearer now ;-)

kenny1999 Feb26-12 10:57 AM

Re: Further Premutation / Combination problem
 
Quote:

Quote by drawar (Post 3784694)
This is called permutation with repetition.
Take 'APPLE' for example, the word contains 2 identical P's but we can distinguish them by adding suffixes, i.e. A(P1)(P2)LE.
Then the number of permutation of the letters A, P1, P2, L, E is 5!
But this number includes separately the 2 permutations A(P1)(P2)LE and A(P2)(P1)LE, so that the arrangement APPLE is counted twice in the 5! permutations.
Because P1 and P2 can be arranged in 2! ways, every distinct arrangement of the letters A, P1, P2, L, E is included 2! in the 5! permutations.
Therefore the numbers of arrangements of the letters A, P, P, L, E is 5!/2!
Things should be clearer now ;-)

oh god, i finally get it.

by the way, i think the formula and calculation is pretty easy on this topic - Permutation and Combination. I always get a big headache when it comes to solving problems on this topic, but once the solution is there, it seems very easy. I am perfect with other Maths topic, i wish to know if there are any technique or better learning materials on the web about this topic. I am very weak at this.


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