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If you have m students, m places, there are m! permutations. Take a certain number of them, k. There are k! positions for them in the first k positions in the line, therefore we can associate k!(m-k)! permutations of the m! initial with that number k. However, there are two other things stated, and I have no idea why: -No different choice for k will arrive at the same permutation -Add up all the permutations for all possible ks, then you achieve m! permutations
1.) If m=6, surely k=2 and k=4 would have the same number o permuations 2.) 6! does not= 5!1!+4!2!+3!3!+2!4!+1!5!
*I don't know, but bookmarked because I want to learn more about combinatorics
How No. 2's actually stated: By considering all possible selections of k students from m we obtain, under the above association, al possible permutations for the m students (the assosiation= k assosiates with k!(m-k)! permutations)
@TuringTest, 'A background to Pascal's triangle' is where this came from, if it helps
For this statement -No different choice for k will arrive at the same permutation I interpret this to mean an actual permutation and not the number of permutations. If m=6, surely k=2 and k=4 would have the same number of permutations k=2 has only 2 members in each permutation, and k=4 has 4 members in each permutation. It is true that each set has the same number of permutations. This By considering all possible selections of k students from m we obtain, under the above association, al possible permutations for the m students (the assosiation= k assosiates with k!(m-k)! permutations) Though it is as clear as mud, it probably means (m choose k) * k!(m-k)! = m! which is true.