Ways of making 10 letter combinations out of 3 given letters Find the number of 10 letter combinations from ${a, b, c}$ that contain:

At least one of each letter
At least two of each letter
At least one A, two Bs, three Cs
Any number of each letter

(Also repetition is allowed)
For number 1 I think that the answer is $\binom{9}{2}$, since there's n-1 places that I can place a bar from stars&amp;bars.
For number 2 though I'm not sure how I can apply this in order to place the bars so that I get at least 2 of each element $a,b,c$.
Edit 2: I did some thinking about question 2 and I think I might have an answer which is correct. Since I need at least 2 of each letter I decided to group 1 of $AA, BB, CC$ so that $AA$ was one object meaning that now in total I had 7 objects. Which then means that I would have $\binom{6}{2}$ places to divide this group up. Please let me know if this approach is correct. Or if I've made a mistake. Thank you.
Let frequency of $a,b,c$ be $x,y,z$ respectively. You're looking for integral solutions of following cases :

$x \ge 1, y \ge 1, z \ge 1$ and $x+y+z=10$
$x \ge 2, y \ge 2, z \ge 2$ and $x+y+z=10$

This is same as $u \ge 1, v \ge 1, w \ge 1$ and $u+v+w=7$

$x \ge 1, y \ge 2, z \ge 3$ and $x+y+z=10$

This is same as $x \ge 1, p \ge 1, q \ge 1$ and $x+p+q=7$

$x \ge 0, y \ge 0, z \ge 0$ and $x+y+z=10$


Hence answers would be $\binom{9}{2}, \binom{6}{2},\binom{6}{2}, \binom{12}{2}$ respectively.
For (1), start with &quot;abc&quot; so that you have fulfilled  the &quot;at least one of each letter&quot; requirement.  You have 7 letters that can be any of the three &quot;a&quot;, &quot;b&quot;, or &quot;c&quot; so there are 3^7 possible choices.  For each such 10 letter combination there are 10! possible arrangements of the letters.  There are 10!(3^7) such combinations.
For (2), start with &quot;aabbcc&quot;.  There are then 4 letters left so 3^4 possible choices.  Since there are again 10! different arrangements, there are 10!(3^4) such combinations.
For (3), start with &quot;abbccc&quot; and finish in the same way.
For (4), with just 10 letters chosen from &quot;a&quot;, &quot;b&quot;, and &quot;c&quot;, there are 3^10 such combinations.