Permutation of the string Ronald Mcdonald I am checking to see if I am on the right track for my Discrete Math class. 
Given the string: RONALDMCDONALD
Vowels: AAOO
How many permutations do NOT have consecutive vowels in them?
I first visualized the problem: _R_N_L_D_M_C_D_N_L_D_
I answered this question by taking the total number of choices: ${11 \choose 4}$
Then the total number of vowel arrangements: ${11 \choose 2}$
Finally, the total number of constant arrangements $\frac{10!}{2!2!3!}$
I multiplied these together for a final answer of:  ${11 \choose 4}$ * ${11 \choose 2}$ * $\frac{10!}{2!2!3!}$
Am I right to assume the total number of vowel arrangements is equal to 11 choose 2?
How many permutations contain the vowels in order (A's before O's)?
Would the answer to this problem be ${11 \choose 4}$?
How many permutations contain the substring OLAND?
If we take OLAND as a super-letter then the letters we have left to permutate are
RNDMCDOAL 
Within this string we have 1 R, 1 N, 2 D, 1 M, 1 C, 1 O, 1 A, 1 L
Therefore the answer is $\frac{10!}{2!}$
 A: 
How many permutations of the string RONALDMCDONALD do not have consecutive vowels in them?

The ten consonants consist of $1$ R, $2$ Ns, $2$ Ls, $3$ Ds, $1$ M, $1$ C.  The four vowels consist of $2$ As and $2$ Os.  
Your idea of arranging the consonants, then separating the vowels by placing them between the consonants or at the ends of the row is sound.  Let's correct your count.
We arrange the ten consonants first.  We have ten positions to fill.  Choose three of them for the Ds, 2 of the remaining seven positions for the Ls, and two of the remaining five positions for the Ls.  This leaves us with three positions for three distinct consonants.  The C, R, and M can be placed in those positions in $3!$ ways.  Hence, the number of arrangements of the ten consonants is 
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! = \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{2!3!} \cdot 3! = \frac{10!}{3!2!2!}$$
As you observed, arranging the ten consonants creates eleven spaces, nine between successive consonants and two at the ends of the row.  To separate the vowels, we must select four of these eleven spaces in which to place a vowel, as you did.  We must then select which two of these four spaces will be filled with As.  The remaining two must be filled with Os.  Hence, the number of ways of selecting spaces for the vowels and arranging them in the selected spaces is 
$$\binom{11}{4}\binom{4}{2} = \frac{11!}{7!4!} \cdot \frac{4!}{2!2!} = \frac{11!}{7!2!2!}$$
The factor of $7!$ in the denominator results from the fact that the seven spaces that are left blank spaces are indistinguishable.  The factors of $2!$ in the denominator represent the fact that the two As are indistinguishable and the two Os are indistinguishable.
Hence, the number of permutations of the string RONALDMCDONALD that do not have consecutive vowels in them is 
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! \cdot \binom{11}{4}\binom{4}{2}$$ 

How many permutations of the string RONALDMCDONALD contain the vowels in alphabetical order?

If we ignore the restriction, we have $14$ letters to arrange.  Of these, we have $1$ R, $2$ Ns, $2$ Ls, $3$ Ds, $1$ M, $1$ C, $2$ As, and $2$ Os.  They can be arranged in 
$$\binom{14}{3}\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3! = \frac{14!}{3!2!2!2!2!}$$
distinguishable ways.  
The four vowels can be arranged in $\binom{4}{2} = 6$ distinguishable ways. Of these six ways, only one leaves the vowels in alphabetical order.  By symmetry, the number of arrangements of the string RONALDMCDONALD in which the vowels appear in alphabetical order is 
$$\frac{1}{6}\binom{14}{3}\binom{11}{2}\binom{7}{2}\binom{5}{2}3!$$

How many permutations of the string RONALDMCDONALD contain the substring OLAND?

If we treat OLAND as a block, we have ten objects to arrange.  They include OLAND, $1$ R, $1$ N, $1$ L, $2$ Ds, $1$ M, $1$ C, $1$ A, $1$ O.  Choose two of the ten positions for the two Ds.  The remaining eight distinct objects can be arranged in the remaining eight places in $8!$ orders.  This gives a preliminary count of 
$$\binom{10}{2}8! = \frac{10!}{2!}$$
which is what you have obtained.
However, notice that it is possible to have two substrings of the form OLAND.  We have counted these strings twice, once for each way we could have designated one of the substrings as the OLAND in the string.  We only want to count them once, so we must subtract them from the total.
If we have two blocks of OLANDs, we also have $1$ R, $1$ D, $1$ M, and $1$ C.  Thus, we have six objects to arrange, including two identical blocks and four distinct letters.  We choose two of the six positions for the blocks, then arrange the four distinct letters in the remaining four positions, which can be done in 
$$\binom{6}{2}4! = \frac{6!}{2!}$$
distinguishable ways.
Hence, the number of permutations of the string RONALDMCDONALD that contain the string OLAND is 
$$\binom{10}{2}8! - \binom{6}{2}4!$$
