Mastering Combinatorics: Exploring the Formula for Permutations

In summary, the conversation revolved around finding the explicit formula for the number of permutations of a given set of elements with specific quantities, as well as the need to elaborate on the solution provided by the student. The student's answer was correct, but it was deemed too brief and lacking in detail. The instructor recommended seeking further clarification and explanation from the instructor for improvement.
  • #1
12john
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Thread moved from the technical forums to the schoolwork forums
My combinatorics professor has a MA, PhD from Princeton University. On our test, she asked

What's the explicit formula for the number of ##p## permutations of ##t## things with ##k## kinds, where ##n_1, n_2, n_3, \cdots , n_k## = the number of each kind of thing ?

I handwrote, but transcribed in Latex, my answer below.

To deduce the formula for all the unique permutations of length ##l## of ##\{n_1,n_2,...,n_k\}##, we must find all combinations ##C=\{c_1,c_2,...,c_k\}## where ##0 \leq c_k \leq n_k##, such that
##\sum_{i=1}^k c_i=l##.

What we need, is actually the product of the factorials of the elements of that combination:
##{\prod_{i=1}^k c_i!}##

Presuppose that the number of combinations is J. Then to answer your question, the number of permutations is
$$= \sum_{j=1}^J \frac{l!}{\prod_{i=1}^k c_i!}
= \sum_{c_1+c_2+...+c_k=l} \binom{l}{c_1,c_2, \cdots ,c_n},$$
as a closed form expression with a Multinomial Coefficient. *QED.*

How can I improve this? What else should I've written? Professor awarded me merely 50%. She wrote
Your answer is correct, but your solution is too snippy. You need to elaborate.
 
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  • #2
12john said:
I handwrote, but transcribed in Latex, my answer below.
I edited your LaTeX. At this site we use MathJax, which has to be delimited by either pairs of # characters (for inline TeX) or pairs of $ characters (standalone).
I also removed all the color stuff. We prefer that you use a minimum of extra color, bolding, italics, etc.
12john said:
How can I improve this? What else should I've written?
What she wrote was "You need to elaborate." The best explanation would come from your instructor.
 

FAQ: Mastering Combinatorics: Exploring the Formula for Permutations

What is the basic formula for calculating permutations?

The basic formula for calculating permutations is given by P(n, k) = n! / (n - k)!, where n is the total number of items to choose from, k is the number of items to arrange, and "!" denotes factorial, which is the product of all positive integers up to that number.

How do permutations differ from combinations?

Permutations consider the order of the items, meaning that different arrangements of the same items are counted as distinct. Combinations, on the other hand, do not consider the order, so different arrangements of the same items are counted as the same.

Can you provide an example of a permutation problem?

Sure! Suppose you have 5 books and you want to arrange 3 of them on a shelf. The number of ways to do this is given by P(5, 3) = 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 60.

What is the significance of factorial in permutations?

Factorial represents the number of ways to arrange a set of items. For a set of n items, there are n! (n factorial) ways to arrange them. In permutations, factorial helps account for the total number of possible arrangements by considering the order of selection.

Are there any special cases or exceptions in permutation calculations?

Yes, one special case is when k = n, meaning you are arranging all items. In this case, the formula simplifies to P(n, n) = n!, which is simply the number of ways to arrange all n items. Another special case is when k = 0, where P(n, 0) = 1, indicating there is exactly one way to arrange zero items.

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