Probability Question please help with the second part

Question

How many different arrangements can be made with the letters in the word DAUGHTER,I got the answer 40320 but don't know how to do the second part How many of these arrangements can be made if the first letter must be a vowel and the last letter must be D

Answer 1

1) The letters are all distinct, so $8!$

2) You have A or U or E on the first position, D on the last position, and 6 other that can be in any order, so $3\cdot 6!$

Answer 2

If we have a word with $n$ letters that are all distinct letters, there is $n!$ possibilities: $n$ choices for the first letter, then $n-1$ for the second letter, then $n-2$ for the third, etc until we have only one letter left.

You seem to have understood that since you obtained $8! = 40320$ possible choices, which is the correct answer.

If you don't know the notation, $n!=n(n-1)\cdots 2\cdot 1$. Example: $4!=4\cdot3\cdot2\cdot1= 24$.

DAUGHTER has three different vowels, hence three possible choices for the first letter. There are $8$ letters in the word, but the last one is forced and the first one has to be one of the three vowels. It remains $8-2=6$ choices.

$$\underbrace{\text{Vowel}}_{3} \cdot \underbrace{\text{six letters}}_{6!}\cdot\underbrace{\text{the letter D}}_{1}=3\cdot 6! \cdot 1 = 3\cdot 720 = 2160$$

As a follow-up question, as you tagged "Probability", you may wonder what is the probability that a randomly chosen word of $8$ letters taken from the word DAUGHTER begins with a vowel and ends by the letter $D$. Without further assumptions, it is customary to consider that the events have equiprobability, which means that the probability to get a particular arrangement is the same for any arrangement. In the case of equiprobable events, we know that:

\begin{align*} &\text{Pr}\left(\text{the word begins with a vowel and end by D}\right)\\ &=\frac{\text{number of possible words beginning with a vowel and ending by D}}{\text{number of possible words}}\\ &=\frac{2160}{40320} \end{align*}

However, in such exercises, it is often preferable to keep the short notations and not do all the calculations, for two reasons. The first being that it is often clearer to understand how one obtained this number ($8!$ refers to the permutation of $8$ distinct elements, for example). The second one is that one can often simplify these short notations at the end. Here, we have:

$$\frac{2160}{40320}=\frac{3\cdot 6!}{8!}=\frac{3\cdot 6!}{(8\cdot 7)\cdot 6!}=\frac{3}{56}\approx 0,0536$$

Answer 3

If the first letter must be a vowel than it has just 3 possible letters, A or U or E, then you've got to put that the first option will be of 3 possible ways, i.e, look for what you did first:

\begin{equation} \_ \cdot \_ \cdot \_ \cdot \_ \cdot \_ \cdot \_ \cdot \_ \cdot \_ \end{equation}

In each one of the spaces you putted a possible letter of the eight you have, then you got the answer $8! = 40320$. Now the first can have just 3 possible letter's, so put the number 3 first, but because the last one have to be D you are left with, for the second letter, just 6 possibilities, then you will have the construction in $3 \cdot (8-2)\cdot (8-3)\cdot (8-4)\cdot (8-5)\cdot (8-6)\cdot (8-7)\cdot 1$ the last possible word have to be D. Then you can construct $2160$ diferent word's.