Number of combinations and permutations of letters

I am confused by the following exercise.

Exercise. Find the number of (a) combinations and (b) permutations of 4 letters each that can be made from the letters of
the word Tennessee.

Ideas:
We have:

T-1

E-4

N-2

S-2

If we would have unlimited number of letters, the number of arrangements is $4*4*4*4$, however the number of T’s, N’s, S’s doesn’t all us to do so, and I don’t see the easy way to eliminate impossible cases.

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You have to be aware of the differences between combinations and permutations.

The first are all the posible agrupations that you can make with the elements of your set, here the order doesn’t matter, $(123)$ is the same as $(312)$, but it’s important that the element doesn’t repeat.

The latter are all the differente agrupations of the elements of your set where the order matter, it’s not the same to say $(123)$ than $(312)$, but you can’t repeat elements, say $(111)$.

First the permutations. You can take the set that cointains the letters from $Tenessee$ like $\{T_1\;E_2\;N_3\;E_4\;S_5\;S_6\;E_7\;E_8\}$, and the size of the sample as 4, imagine that you have 4 boxes, in each box you can put those elements, but with the restriction that you can’t repeat the elements, and the order matters,

$\mathbf (Box 1)$ you can put any of the 8 elements

$\mathbf (Box 2)$ since you already put one element in B1 and you can’t repeat, you have 7 choices

$\mathbf (Box 3)$ using the same thinking, now you have 6 choices

$\mathbf (Box 4)$ now you have 5 choices

So using the Rule of product, you have that the permutations that you can get are $8*7*6*5$

Now, with the combinations, you want to arrange the 8 elements in 4 boxes again, but you’ll consider cases like $\{T_1\;E_2\;N_3\;E_4\}$, $\{E_2\;T_1\;N_3\;E_4\}$, $\{E_2\;N_3\;T_1\;E_4\}$, $\{E_2\;N_3\;E_4\:T_1 \}$, etc. to be the same. In the case of permutations, these elements were different, so will have to discard this cases if we want to get combinations, so the number of extra elements is $4*3*2*1=4!$ (see the EDIT), so the combinations are $$8*7*6*5 \over 4*3*2*1$$

You could also use formulas. If $n$ is the number of elements in your set, and $m$ is the size of your sample, then the permutation is given by: $$n! \over (n-m)!$$ There are books that use the term variation for this formula, and the term permutation for $n!$, but you can see that it’s the same. And the combinations are given by: $$n! \over m!(n-m)!$$ wich is the definition of binomial coefficient

In your excersize $n=8$ and $m=4$.


$\mathbf {EDIT:}$ Why there are $4!$ extra cases?

I think it’ll be easier with a smaller example first.

There’s a group of 5 students that want to form a committee of 3, how many different committees can they form?

Using the same box method, we have 3 spots, we know that the first spot can be taken by 5 students, the second by 4 students and the third by 3 students, now we are not done yet, because by now we are considering that the committee of $(A,B,C)$ is different from $(B,A,C),(B,C,A),(A,C,B),(C,A,B),(C,B,A)$ (wich as you can see, is the permutations of 3 elements in 3 spots, hence $3*2*1=3!$), our interest is to know how many times are we counting the same committee, so let’s consider one fixed committee, like the one we had $(A,B,C)$, and count how many times we can order those elements, we know that: $3!$; now lets fixed another one: $(B,C,D)$ we are also counting $3!$ more times, another one $(C,D,E)$ the same, etc. etc. The same for the rest. Then for every 3 people we are counting $3!$ more times the product $5*4*3$. So the answer is $$5*4*3 \over 3!$$
Now with your problem, I took one fixed posiblity $\{T_1\;E_2\;N_3\;E_4\}$, and saw that you can arrange those elements $4!$, using the same thinking as the problem above, what we are doing is counting $4!$ times more the product $8*7*6*5$, hence what we are looking for is $$8*7*6*5 \over 4!$$
Hope this helps. You can always make drawings to see what happens, try it with the committee’s problem.

Multisets of size $4$ made from the letters in TENNESSEE can have frequencies of $T$, $N$ and $S$ as per the following table; all of the remaining letters must be E.

$$\begin{array}{c|ccc|c}
\text{number} & \# \text{T} & \# \text{N} & \# \text{S} & \text{multiset} \\
\hline
1 & 0 & 0 & 0 & \{\text{E,E,E,E}\} \\
2 & 0 & 0 & 1 & \{\text{S,E,E,E}\} \\
3 & 0 & 0 & 2 & \{\text{S,S,E,E}\} \\
4 & 0 & 1 & 0 & \{\text{N,E,E,E}\} \\
5 & 0 & 1 & 1 & \{\text{N,S,E,E}\} \\
6 & 0 & 1 & 2 & \{\text{N,S,S,E}\} \\
7 & 0 & 2 & 0 & \{\text{N,N,E,E}\} \\
8 & 0 & 2 & 1 & \{\text{N,N,S,E}\} \\
9 & 0 & 2 & 2 & \{\text{N,N,S,S}\} \\
10 & 1 & 0 & 0 & \{\text{T,E,E,E}\} \\
11 & 1 & 0 & 1 & \{\text{T,S,E,E}\} \\
12 & 1 & 0 & 2 & \{\text{T,S,S,E}\} \\
13 & 1 & 1 & 0 & \{\text{T,N,E,E}\} \\
14 & 1 & 1 & 1 & \{\text{T,N,S,E}\} \\
15 & 1 & 1 & 2 & \{\text{T,N,S,S}\} \\
16 & 1 & 2 & 0 & \{\text{T,N,N,E}\} \\
17 & 1 & 2 & 1 & \{\text{T,N,N,S}\} \\
\end{array}$$

For the permutation problem, we just sum the corresponding multinomial coefficients for the multisets in the above table.