anonymous
  • anonymous
nCk= The combinations of n taken k at a time. which of the following is equal to 11C8? a) 11!/3! b) 11!/11!8! c) 11!/8!3! d) 11!/8!
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Is it a..?
anonymous
  • anonymous
no c
anonymous
  • anonymous
nCr = n!/r!(n-r)!

Looking for something else?

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

More answers

anonymous
  • anonymous
c is the right choice
anonymous
  • anonymous
nPk = n!/(n-k)1
anonymous
  • anonymous
right
anonymous
  • anonymous
dont be confused between permutations and combination , both are different
anonymous
  • anonymous
yep agreed :D
anonymous
  • anonymous
I'm confused between the two formulas and two answers
anonymous
  • anonymous
11C8= 11! ----- 8!*3!
anonymous
  • anonymous
i have a way combination : we only select the items permutation: we select and we also arrange the items
anonymous
  • anonymous
umm .. sorry that will be when u do problems of probability
anonymous
  • anonymous
but metarock ur definition is right
anonymous
  • anonymous
ya but chizzle was confused in the formulas , a different context though
anonymous
  • anonymous
Ah, just found a way to remember (P)ermutation ... (P)osition matters both begin with P
anonymous
  • anonymous
(C)ombination and (C)ounting (or selecting) ;-)
anonymous
  • anonymous
So then it is 11!/8!3! ?
anonymous
  • anonymous
yups
anonymous
  • anonymous
Okay thank you all for the help/formulas/and differences between them
anonymous
  • anonymous
ur welcome

Looking for something else?

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