anonymous
  • anonymous
A binomial query
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
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!
TuringTest
  • TuringTest
*I don't know, but bookmarked because I want to learn more about combinatorics

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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)
anonymous
  • anonymous
@TuringTest, 'A background to Pascal's triangle' is where this came from, if it helps
phi
  • phi
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.