A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
how many bit strings length 9 contain exactly 3 ones?
anonymous
 5 years ago
how many bit strings length 9 contain exactly 3 ones?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a combinatorics problem: another way to state this would be: how many different ways can you chose 3 things out of 9? This is often written 9C3 or "9 choose 3" and the formula for nCk is \[\frac{n!}{k!(nk)!}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would the answer be 42?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you come to that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i plugged in 9 for n and 3 for k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then did 9*8*7 divided by 3*2*1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't forget the (nk)! part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i'm sorry, you're right, that should give you the right thing, but that gives you 84 i think, not 42

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now it says show that c(9,3) = c(9,6) by describing a matching of each bit string of length 9 with 3 ones with a bit string of length 9 with 6 ones

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you choose 3 things from 9, you can think of that as partitioning your original set into two groups, one with 3 items and one with 93=6 items. In the case of this problem, one group was the set of positions that have a 1, and the other group was the set of positions that have a zero. So really, the question isn't "how many ways are there to choose 3 from 9", but more accurately "how many ways can I partition 9 things into a group of 3 and a group of 6." When you think of it in these terms, what you assign to the groups is completely arbitrary. So the number of ways to assign 3 ones and 6 zeros is exactly the same as the number of ways to assign 6 ones and 3 zeros.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can check this intuition with the formula: notice that the formula doesn't care whether you plug in 9C3 or 9C6, since the two terms in the denominator just swap.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.