Combinatorics: How Many Ways Are to Arrange the Letters in VISITING?

In summary, there are a total of 5! ways to arrange the letters of "VISITING" without any consecutive I's.
  • #1
Shoney45
68
0

Homework Statement



How many ways are there to arrange the letters of the word VISITING with no pairof consecutive I's?

Homework Equations



C(n,k) P(n,k)

The Attempt at a Solution



I am calculating the entire number of arrangements possible at P(8,5). I then want to find out the entire number of cases where there are at least two I's consecutively placed, and subract that from P(8,5).

Since there are eight places to put letters, there are seven possible ways to places to I's consecutively. What I am unsure of is if those seven arrangements of letters include the six possible cases where all three I's are consecutively placed.
 
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  • #2
Based on the problem wording I would defintely consider cases where all three I's are consecutive - in fact exclusing these woul make the problem more difficult

I thin you could probably do this with cases without much difficulty

consider
IIxxxxxx
now there is 6! ways to choose the remaining letters, it doesn't matter where you put the 3rd I as you already have two consecutive Is.

now consider
xIIxxxxx
the only tricky part will be when you choose IIIxxxxxx, which has already been counted, so there is only 5.5! other ways to choose the other 6 letters

hopefully you can continue with this, note that I've assumed the Is are indistinguishable
so VISITING counts as one way to arrange the letters, even though were the Is distinguishable (or say teh letters were drawn at random) they could be arranged 3! different ways
 
  • #3
lanedance said:
Based on the problem wording I would defintely consider cases where all three I's are consecutive - in fact exclusing these woul make the problem more difficult

I thin you could probably do this with cases without much difficulty

consider
IIxxxxxx
now there is 6! ways to choose the remaining letters, it doesn't matter where you put the 3rd I as you already have two consecutive Is.

now consider
xIIxxxxx
the only tricky part will be when you choose IIIxxxxxx, which has already been counted, so there is only 5.5! other ways to choose the other 6 letters

hopefully you can continue with this, note that I've assumed the Is are indistinguishable
so VISITING counts as one way to arrange the letters, even though were the Is distinguishable (or say teh letters were drawn at random) they could be arranged 3! different ways

I think that I would count it by first determining how many ways you can arrange the 5 non-I's. There are 5! ways to do this. Now, look at the "diagram" below. The $'s represent non-I's and the %'s represent spaces between them:

%$%$%$%$%$%

Each of the percent symbols are valid locations for ONE I. Thus, for each of the 5! permutations of the non-I's, we just need to determine how many ways there are to place I's so that there are no to back to back. This is exactly the same as pick three of the percent signs. Since there is C(6,3) ways to do that, there must be C(6,3) ways to place the I's. So, if I have thought this through correctly, there are 5!C(6,3) ways to arrange the letters VISITING subject to the given constraints.
 
  • #4
yep that works well
 
  • #5
Thanks you guys. That was all very helpful.
 

1. How many different ways can the letters in VISITING be arranged?

There are 8 letters in the word VISITING, so the total number of arrangements would be 8!. This can be calculated as 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320.

2. How many arrangements are possible if the letters are arranged in alphabetical order?

If the letters in VISITING are arranged in alphabetical order, there would be only one possible arrangement, as it is already in alphabetical order.

3. How many arrangements are possible if the vowels (I, I, and I) are kept together?

There are 3 vowels (I, I, and I) in VISITING, so they can be treated as one group. Therefore, there would be 7 letters to arrange, which can be done in 7! ways. However, since there are three identical I's, we need to divide by 3! to account for the duplicates. This gives us a total of 7!/3! = 840 possible arrangements.

4. How many arrangements are possible if the letters G and T are always next to each other?

If G and T are always next to each other, we can treat them as one group. This gives us 7 letters to arrange, which can be done in 7! ways. However, since there are two identical G's, we need to divide by 2! to account for the duplicates. This gives us a total of 7!/2! = 3,360 possible arrangements.

5. How many arrangements are possible if the letters S and N are never next to each other?

If S and N are never next to each other, we can treat them as two separate groups. This gives us 6 letters to arrange, with 3 letters in each group. This can be done in 6!/3! = 120 ways. However, since there are two identical I's, we need to divide by 2! to account for the duplicates. This gives us a total of 120/2! = 60 possible arrangements.

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