1. combination letters with restrictions

How many arrangements are there for PROBABILITIES if there are

- no 2 vowels together

- there is no E next to an I

2. b) These are the consonants are: P,R,B,B,L,T,S.
They can be arranged in $\displaystyle \frac{7!}{(2!)}$ ways;
Each of the ways creates 8 places _P_R_B_B_L_T_S_ to put the vowels.
The number of places is then $\displaystyle 8 \choose 6$. The number of vowel arrangements is $\displaystyle \frac{6!}{(3!)}$
Put it together: $\displaystyle {\frac{7!}{(2!)}} {8 \choose 6}$$\displaystyle \frac{6!}{(3!)}$.

a) “PROBABILITIES” the word has 13 letters but some repeat.
There are $\displaystyle \frac{13!}{(3!)(2!)}$ ways to arrange this word: 3 I’s & 2 B’s.
You now must find the number of ways we can have “EI”, “IE”, or “IEI”
And subtract from total.

3. Arrangements with repeated letters

Hello PatCal
Originally Posted by PatCal
How many arrangements are there for PROBABILITIES if there are

- no 2 vowels together

- there is no E next to an I
The standard method for solving letter re-arrangements is to assume first that any repeated letters are distinguishable - so here we would call the two B's $\displaystyle B_1$ and $\displaystyle B_2$, and the three I's $\displaystyle I_1$, $\displaystyle I_2$ and $\displaystyle I_3$. Then work out the number of arrangements with all the letters different. Here that would be 13!.

Finally note that because of repeated letters, there will be duplications. So we divide this total by the number of arrangements of each group of repeated letters within themselves. In the case of the 2 B's, that's 2!, and for the 3 I's, it's 3!. So the answer would be

$\displaystyle \frac{13!}{2!3!}$

But we have extra restrictions, so we shall need to apply additional techniques.

So, for question 1, where no two vowels are together. There are six vowels: O, A, E and three I's; and seven consonants. If no two vowels can come together, and we number the positions 1 to 13, then the vowels must occupy positions 2, 4, 6, 8, 10 and 12; and the consonants the odd-numbered positions.

So, the six vowels could be arranged in their positions in 6! ways if they were all different. But, because of the duplication among the three I's we need to divide this number by 3!, the number of ways of arranging the I's among themselves. Thus the number of ways in which the vowels can be placed is

$\displaystyle \frac{6!}{3!}$

Now do the consonants in a similar way, remembering to divide to take into account the duplication of the two B's.

Finally, multiply the number of arrangements of the vowels by the number of arrangements of the consonants to get the total.

For question 2, where there is no E next to an I. This is most easily solved by finding the number of arrangements in which an E and an I will come together and subtracting this from the overall total. So, how many ways are there of combining an I and an E into a single letter-group? Clearly, there are just two: IE or EI.

Now treat this letter group as a single character, and work out how many ways there now are of arranging this in a line together with the other eleven letters, remembering to take into account the duplication that will arise because of the 2 B's and the remaining 2 I's. *See PS

Subtract this from $\displaystyle \frac{13!}{2!3!}$ and you're done.

Can you do it from here?

*PS. Sorry, there's an additional factor to take into account. That is, the number of times the arrangement IEI occurs, because each of these will be counted twice in the above method: once where the IE group is placed in the line, and then has an I placed on its right, and once with the EI group being placed and then an I on its left.

So, you need to work out how many times you'll get the combination IEI occurring, and add this back in - because we shall have taken it away twice, when once would have been correct. This number, then, is the number of arrangements of the letter-group IEI and the remaining ten letters - which this time only have B as a duplicated letter. Number of arrangements containing IEI, then, is

$\displaystyle \frac{11!}{2!}$

and I make the final answer to question 2: 299,376,000

Tricky!

4. Hello, PatCal!

I think I have the first part.
(I'm still working on the second part.)

How many arrangements are there for PROBABILITIES if there are

(a) no two vowels together

$\displaystyle \text{We have: }\:\underbrace{\{A,E,I,I,I,O\}}_{\text{6 vowels}} \cup \underbrace{\{B,B,L,P,R,S,T\}}_{\text{7 consonants}}$

Place the 7 consonants in a row, leaving a space before, after, and between them.

. . $\displaystyle \begin{array}{ccccccccccccccc}\_ & C & \_ & C & \_ & C & \_ & C & \_ & C & \_ & C & \_ & C & \_ \end{array}$

There are: .$\displaystyle \frac{7!}{2!}$ orderings for the consonants.

Place the 6 vowels in the spaces.
. . There are: .$\displaystyle {8\choose6}$ choices for the six spaces.
. . and there are: .$\displaystyle \frac{6!}{3!}$ orderings for the vowels.

Therefore: .$\displaystyle \frac{7!}{2!}\cdot{8\choose6}\cdot \frac{6!}{3!} \:=\:2520 \cdot 28 \cdot 120 \:=\:8,\!467,\!200$

5. Arrangements with repeated letters

Hello PatCal and Soroban -

My first posting was incorrect. My apologies. In question 1, the vowels don't have to be placed in positions 2, 4, 6, 8, 10 and 12, of course. They could go in any six of the eight spaces, as Soroban has said!