Number of ways to rearrange in the word UNSUCCESSFULLY with the given requirements? Question: Given the word UNSUCCESSFULLY, calculate the number of ways to rearrange word with the given requirements below:

*

*Letter 'Y' appears after all the vowels in the word?

*The first 'S' appears before the first 'U'?

*Both U's and L's appear consecutively at the same time?

*C's don't appear consecutively?

Note: Each question is separate from the others.
My attempted solution:
For question 1, I placed the 4 vowels and the letter 'Y' first and then rearranged the word and got:
$${\binom{14}{5}} (\frac{4!}{3!})(\frac{9!}{2!2!3!3!})$$
For question 2, I did the exact same and got
$${\binom{14}{6}} (\frac{4!}{2!2!})(\frac{8!}{2!2!})$$
I'm honstly a bit confused of what the difference is between question 1 and 2. I might not be right but I used the same approach in the two answers I got above.
For question 3, I bascially combined the 2 L's and also the 3 U's and treaten them as one single letter. This resulted in the word having 11 letters. And then I got the answer:
$${\binom{11}{3}}{\binom{8}{2}}{\binom{6}{1}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{1}}{\binom{2}{1}}{\binom{1}{1}}$$
For question 4, I caclculated the max possible ways to rearrange the letter and subtracted the ways at which three C's are together from that. I got:
$${\binom{14}{3}}{\binom{11}{1}}{\binom{10}{3}}{\binom{7}{2}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{2}}{\binom{1}{1}} - {\binom{13}{3}}{\binom{10}{1}}{\binom{9}{3}}{\binom{6}{1}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{2}}{\binom{1}{1}}$$
I all honesty, I only followed the steps to which I think it might be correct and most of these are likely to be incorrect. I'd appreciate any help!
 A: Total number of combinations of UNSUCCESSFULLY:
> factorial(14)/1/1/1/2/2/6/1/6
[1] 605404800

Answers:

*

*First see how many combinations there are of UEUUY where Y appears last. There are 4. Choose 5 slots for it in the 14 slots. Then, rearrange the remaining 9 letters anyhow you want.

$$4*{14\choose 5}*\begin{pmatrix}9\\1,1,2,2,3\end{pmatrix}$$
> 4*choose(14,5)*factorial(9)/1/1/2/2/6
[1] 121080960
This is 20%.



*See how many combinations of SSSUUU with the first S appearing before the first U. The possibilities are SU_ _ _ _ , SSU _ _ _ and SSSUUU. Choose 6 slots for the S's and U's. Fill in the remaining 8 slots any way you want.

$$\left(\begin{pmatrix}4\\2\end{pmatrix}+3+1\right)*\begin{pmatrix}14\\6\end{pmatrix}*\begin{pmatrix}8\\1,1,1,1,2,2\end{pmatrix}$$
> (factorial(4)/2/+3+1)*choose(14,6)*factorial(8)/2/2
[1] 151351200
This is 25%.



*I assume you meant all 3 U's appear consecutively and all 2 S's. Treating the 3 U's as one block and similarly the 2 S's, the problem reduces to.

$$\begin{pmatrix}11\\1,1,1,1,1,1,2,3\end{pmatrix}$$
> factorial(11)/2/6
[1] 3326400
This is roughly .549%.



*This is easiest done by subtracting the number of times C's do appear consecutively from the total number of arrangements.

$$\begin{pmatrix}14\\1,1,1,1,2,2,3,3\end{pmatrix}-\begin{pmatrix}12\\1,1,1,1,2,3,3,\end{pmatrix}*13$$
> factorial(14)/2/2/6/6-factorial(12)/2/6/6*13
[1] 518918400
This is roughly 85.7%.

A: You made some errors, the reason for which I can hardly explain. So the correct answers:
Q.1
$${\binom{14}{5}} \frac{4!}{1!3!}\frac{9!}{1!1!2!2!\color{red}{3!}}$$
Q.2
$${\binom{14}{6}} \frac{\color{red}{5!}}{2!\color{red}{3!}}\frac{8!}{1!1!1!1!2!2!}$$
Q3 and Q4 are correct but it is possibly better to write:
$$
\frac{11!}{1!1!1!1!1!1!2!3!}.
$$
Note that I write all factors in the denominator. So one can easier check, that the sum of the numbers is equal to the number in numerator.
