+1 vote
339 views
How many ways, can sum be equal to 12 of 3 dice?

Solution:
x1+x2+x3=12
where 1<=x1<=6;1<=x2<=6; 1<=x3<=6
How to solve it further?
retagged | 339 views
Bars and stars approach can also be used, if we restrict one star each in 3 partitions we will have 11C9 combinations i.e 55 . From this we should subtract case when any partition has greater than 7 stars. 7 star in x1 , x2 1 star and x3 1 star, select remaining 3 stars from 3 star+2 bars i.e 5 C 3 .This will be 3 times as we can place 7 stars in x1 or x2 or x3. So 11C9 - 3(5C3) = 25. But I think generating functions process is much easier.

Higher order terms are neglected and only $x^3$ and $x^9$ terms are useful.

selected

you explained very well! much obliged!!
Please let me know how how did you reach this equation
(x^6+x^5+x^4+x^3+x^2+x)3
and why are you loking for the coefficient of x^12, please provide details

it is same like distribution of indistinguishable balls into distinguishable boxes.

Say where x1+x2+x3=12

1<=x1<=6;1<=x2<=6; 1<=x3<=6

x1,x2,x3 can have value between 1,2,3,4,5,6.

Each have the same (x+x^2+x^3+x^4+x^5+x^6) so taken cube ....and n=12 so finding 12th term

http://gateoverflow.in/89753/discrete-maths

@Prabhanjan

on RHS side you it is a expansion $\binom{n+2}{n}x^{n}$

I want $x^{12}$ then so did $x^3 * x^9$ and $-3x^9 * x^3$

coefficient of x^9 is 11C9 (for first ) and x^3  is 5c3.
thanking you in words not enough!! i knew nothing about generating functions. now i am much comfortable with them. Much obliged!!
Welcome  Vijay :)
–1 vote
I guess we can solve it using Generating function Concept
I am getting answer as 70
edited

find coeff of x12 in (x^6+x^5+x^4+x^3+x^2+x)3

@saurabh
can you please tell how did you reach this equation and how to calculate coeffcient of x^12 in it?
I am not that much stronger in maths, hence a detailed solution is required
@vijay i think u should read generating function from rosen book so that u ll get every thing that u want 2 know, it is hardly takes 3-4 hours and after that if u have any doubt u can ask here.:)