<s>[INST] true false combinatorics questions about arrangements I've solved some exercises regarding combinatorics. The questions are true\false questions, but I'm not sure that I've done them right. Please take a look and indicate any corrections I need to make.
1)  The number of different arrangements of the string 'AAABBC' is larger than the number of arrangements of the string 'AABBCC'
2)  The number of different arrangements of the string 'AABBCC' is equal to the number of possible 'cuttings' (subsets) with $3$ classes that include $2$ elements each of the set $\{1,2,3,4,5,6\}$.
3)  The number of different arrangements of the string 'AABBCC' is equal to the number of ways in which $6$ students can submit $3$ different tasks in pairs.
4)  The number of ways in which you can put $3$ identical apples in $4$ different baskets is equal to the number of ways in which you can put $4$ identical apples in $3$ different baskets.
5)  The number of ways in which you can put $4$ different apples in $4$ different baskets is $16$ times the number of ways in which you can put $4$ identical apples in $4$ different baskets.
6)  The number of different separations of the set $a= \{1,2,3,4,5,6\}$ to classes with $3$ elements each is equal to the number of possible combinations of a choosing a set with $3$ elements from $a$.
What I did:
1)  False. If $a$ is the number of As, and $b$ is the number of Bs and $c$ is the number of Cs, then $a = 3$, $b=2$ and $c=1$, and the possible ways to arrange the string'AAABBC' is:$\frac {6!}{3!2!}$. regarding 'AABBCC': $\frac {6!}{2!2!2!}$, which means that the latter is larger.
2)  False. Using the same method, the different arrangements of 'AABBCC' is equal to $\frac{6!}{2!2!2!}$, and the number of possible mappings from $\{1,2,3,4,5,6\}$ to subsets with $3$ classes with $2$ elements each is: $\frac{\binom {6}{2} \binom {4}{2} \binom {2}{2}}{2}$ (I'm sure I've done a mistake here). 
3)  False. Arranging 'AABBCC' is equal to $\frac{6!}{2!2!2!}$, and the number of possibilities in which $6$ students can submit $3$ different tasks in pairs is $\frac{6 \cdot 3}{2}$.
4)  True.  Seems to be equal, since if you have $4$ apples and $3$ baskets, you can only put $3$ apples in the baskets.
5)  False. The number of ways in which you can put $4$ different apples in $4$ different baskets is: $\binom {4+3}{4}$, and the number of ways in which you can put $4$ identical apples in $4$ different baskets is not $16$ times less than that. If I made a mistake in calculation, please correct me if possible.
6)  True. From what I understand, it's basically doing the same.
Please correct me if I've made a mistake, so I can learn and improve. Thank you very much.
 [/INST] 
The number of different arrangements of the string AAABBB is larger than the number of different arrangements of the string AABBCC.

Your work is correct.

The number of different arrangements of the string AABBCC is equal to the number of ways of cutting the set $\{1, 2, 3, 4, 5, 6\}$ into three subsets with two elements each.

You are correct that the number of distinguishable arrangements of AABBCC is 
$$\frac{6!}{2!2!2!}$$
However, your count of the number of ways to divide the set $\{1, 2, 3, 4, 5, 6\}$ into three subsets with two elements each is incorrect.  
Method 1:  The number of ways to divide the set $\{1, 2, 3, 4, 5, 6\}$ into three labeled subsets with two elements each is
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
since there are $\binom{6}{2}$ ways to select two of the six elements for the subset with the first label, $\binom{4}{2}$ ways to select two of the four remaining elements for the subset with the second label, and one way to place the remaining elements in the subset with the third label.  However, the subsets are not labeled. Since the order in which we list the subsets does not matter and there are $3!$ ways we could order the subsets, the number of distinguishable ways of dividing the set $\{1, 2, 3, 4, 5, 6\}$ into three two-element subsets is 
$$\frac{1}{3!}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{1}{6} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{1}{6} \cdot \frac{6!}{2!2!2!}$$
Hence, the statement is still false.
Method 2:  There are five choices for the element that is placed in the same subset as $1$.  That leaves four elements.  There are three ways of placing a number in the same subset as the smallest remaining element.  The remaining two elements must be placed in the third subset.  Hence, there are 
$$5 \cdot 3 \cdot 1$$
ways of dividing the set $\{1, 2, 3, 4, 5, 6\}$ into three two-element subsets.

The number of different arrangements of the string AABBCC is equal to the number of ways in which six students can submit three different tasks in pairs.

There are $\binom{6}{2}$ ways to select which two of the six students submit task 1, $\binom{4}{2}$ ways to select which two of the four remaining students select task 2, and one way for the remaining pair of students to submit task three.  Hence, the number of ways in which six students can submit three different tasks in pairs is 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{2!2!2!}$$
which is equal to the number of distinguishable arrangements of the string 
AABBCC.

The number of ways you can put three identical apples in four different baskets is equal to the number of ways you can put four identical apples in three different baskets.

Let $x_j$ be the number of apples placed in the $j$th basket.  
The number of ways you can put three identical apples in four different baskets is the number of solutions of the equation 
$$x_1 + x_2 + x_3 + x_4 = 3$$
in the nonnegative integers.  A particular solution corresponds to the placement of three addition signs in a row of three ones.  For instance, 
$$+ 1 + 1 1 +$$
corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 2$, and $x_4 = 0$.  There are 
$$\binom{3 + 3}{3} = \binom{6}{3}$$
such solutions since we must choose which three of the six positions for the six symbols (three ones and three addition signs) will be filled with addition signs.
In general, the number of solutions of the equation 
$$x_1 + x_2 + x_3 + \ldots + x_k = n$$ 
in the nonnegative integers is 
$$\binom{n + k - 1}{k - 1}$$
since we must choose which $k - 1$ positions of the $n + k - 1$ positions needed for the $n$ ones and $k - 1$ addition signs will be filled with addition signs.
The number of ways four identical apples can be placed in three different baskets is the number of solutions of the equation
$$x_1 + x_2 + x_3 = 4$$
in the nonnegative integers, which is 
$$\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2}$$
Since $\binom{6}{3} > \binom{6}{2}$, there are more ways to distribute three identical apples to four different baskets than there are ways to distribute four identical apples to three different baskets.

The number of different ways you can put four different apples in four different baskets is $16$ times the number of ways you can put four identical apples in four different baskets.  

As you concluded in the comments, there are $4^4$ ways to put four different apples in four different baskets and $\binom{4 + 4 - 1}{4 - 1} = \binom{7}{4}$ ways to place four identical apples in four different baskets.  Since 
$$4^4 = 256 \neq 16 \cdot 35 = 16\binom{7}{3}$$
the statement is false.

The number of different separations of the set $A = \{1, 2, 3, 4, 5, 6\}$ is equal to the number of ways of choosing a subset of three elements from $A$.  

The statement is false.
There are $$\binom{6}{3}$$ ways to select a subset with three elements from a six-element set.  
We will count the number of ways to separate set $A$ into two sets with three elements in two ways.
Method 1: Since choosing a subset of three elements of $A$ also determines its complement, choosing a subset of three elements counts each separation twice, once when you count the subset and once when you count its complement. Thus, there are 
$$\frac{1}{2}\binom{6}{3}$$
ways to separate set $A$ into two three-element subsets.
Method 2:  In any separation of set $A$ into two sets with three elements, there are $$\binom{5}{2}$$ ways to choose which two of the other five elements will be in the same subset as $1$.  
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