Ways to arrange AABBBCCCDD?

Question

Consider the following question:

How many ways are there to order AABBBCCCDD?

My intuition would be to pick two spots from the 10 for letter A, then 3 spots from 8 for the letter B and so on. So our solution would be (10C2)(8C3)(5C3).

However I saw a solution formatted as the following:

10!/(2!3!3!2!)

However I do not understand how they derive this solution. What is the intuition behind that expression?

Have you tried simplifying $\binom{10}{2}\binom{8}{3}\binom{5}{3}$? — N. F. Taussig, Sep 25 '22 at 09:36
$\binom{10}{2}\binom{8}{3}\binom{5}{3} = \frac{10!}{2!8!} \cdot \frac{8!}{3!5!} \cdot \frac{5!}{3!2!} = \frac{10!}{2!3!3!2!}$. — N. F. Taussig, Sep 26 '22 at 23:54

true blue anil · Answer 1 · 2022-09-25T22:05:02.850

3

They are all forms of multinomial coefficient which you should look up.

$$\binom{10}{2,3,3,2} \equiv \binom{10}2\binom83\binom53\binom22 \equiv \frac{10!}{2!3!3!2!}$$

Although you can use any of the equivalent forms, I like to think of the three forms as:

(i) putting distinct (=labeled) objects into boxes with matching labels
(ii) forming labeled teams from a pool of people
(iii) permuting a word which has some repeating letters

An intuitive understanding of (iii) can be seen in this answer

edited Sep 25 '22 at 22:05

answered Sep 25 '22 at 08:59

true blue anil

48,250

Does the \Large\binom look as wonky to you as it does to me? The parentheses aren't large enough at that size. I wonder why that happens. – Brian Tung Sep 25 '22 at 19:10
@BrianTung: Idk, when i use normal size, on my laptop the figures jump out of the parentheses in the multinomial. (but not on my phone). I am editing it to normal size, could you please let me know if it looks ok. – true blue anil Sep 25 '22 at 19:54
Yeah it looks fine now, though of course it is smaller now. – Brian Tung Sep 25 '22 at 20:29
@BrianTung: Ok, thanks. $;;$ :) – true blue anil Sep 25 '22 at 20:34
Oh, it didn't even occur to me that you weren't using a display mode equation. – Brian Tung Sep 26 '22 at 01:41

score 0 · Answer 2 · answered Sep 25 '22 at 08:58

Instead of a,b,c, and d if we had 10 different numbers we would do 10!. This 10! factorial would be our starting point.

Now let's say you have a set of two numbers, AB and AA AB would have two rearrangements, AB and BA. AA would have one arrangement, AA. When we have a set of numbers and we have values that are the same, they are counted as the same number. Because of this, you would divide the number of ways where the same values could rearrange.

An example is the set ABCDD If we had ABCDE, the number of rearrangements would be 120. If you do 5! for ABCDD, you are counting ABC(D1)(D2) and ABC(D2)(D1) as different arrangements. Because of this, AABBBCCCDD would have 10!/2!(A's) 3!(B's) 3!(C's) 2!(D's).

Ways to arrange AABBBCCCDD?

2 Answers2