## AravindG Group Title find the no: of integral solutions of inequation x+y+z+u<=65 2 years ago 2 years ago

1. FoolForMath Group Title

0 is included?

2. FoolForMath Group Title

I can do this using binomial coefficients.

3. AravindG Group Title

i dont know tht fool

4. FoolForMath Group Title

Using stars and bars, the answer is given by $\huge \sum \limits_{n=0}^{65} \binom{n+3}{3}$

5. satellite73 Group Title

can't wait to see this

6. satellite73 Group Title

stars and bars??

7. AravindG Group Title

hw?

8. FoolForMath Group Title
9. moneybird Group Title

this question is like putting three bookmarks in a book

10. satellite73 Group Title

how cool is that? stars and bars, nice visual!

11. FoolForMath Group Title

Yeah it's coool :D

12. AravindG Group Title

coool

13. shaan_iitk Group Title

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!)

14. FoolForMath Group Title

@shaan_iitk How do you think stars and bars work? That is exactly the same thing you elaborated.

15. shaan_iitk Group Title

means?

16. shaan_iitk Group Title

Sorry but I am not aware of stars and bars ..

17. FoolForMath Group Title

I have posted the url before.

18. shaan_iitk Group Title

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..