Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
AravindG
Group Title
find the no: of integral solutions of inequation x+y+z+u<=65
 2 years ago
 2 years ago
AravindG Group Title
find the no: of integral solutions of inequation x+y+z+u<=65
 2 years ago
 2 years ago

This Question is Closed

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
0 is included?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
I can do this using binomial coefficients.
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.3
i dont know tht fool
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Using stars and bars, the answer is given by \[ \huge \sum \limits_{n=0}^{65} \binom{n+3}{3} \]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
can't wait to see this
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
stars and bars??
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.2
this question is like putting three bookmarks in a book
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
how cool is that? stars and bars, nice visual!
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Yeah it's coool :D
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
One can also think logically and get the solution .. say you have x+y+z+u <= k (some integer) and you have to find the number of solutions then it is like finding the number of ways where k sweets can be distributed to 4 kids and total sweets distributed may not be equal to k. Now lets imagine a situation where we have 4 women and k men. How many ways of arranging them? (4+k)!/(4!*k!) Now say I say that all men who come to the left of any woman are given to that woman (not literally) .. then we have solved our distribution problem and hence our equation.. so I believe the answer should be 69!/(4!*65!)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
@shaan_iitk How do you think stars and bars work? That is exactly the same thing you elaborated.
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
Sorry but I am not aware of stars and bars ..
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
I have posted the url before.
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
okk .. I didn't look at that.. I only looked at your expression which was in summation form.. I thought logically we don't need that summation..
 2 years ago
See more questions >>>
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.