11904 - One Unit Machine

[crip121]

can anyone give me any hint.

[lgarcia]

You can solve it with binomial coefficients and modular multiplicative inverse.

[mubashwir]

@ : crip121

let, there are 4 jobs

Code: Select all

Job no.  number of unit
J1             3
J2             4
J3             5
J4             4

If you find the total number of ways for each Job. Then you can easily find the Total number of Ways for all jobs by multiplying the number of ways of each job.

Now,
i) We need to complete J1 first, So keep one Unit of J1 in the right most area like ____________|J1 . So we can put 2 J1 in the left side in 1 way.

ii) now one current combination is : J1 J1 J1. We need to complete J1 before J2. So we keep one unit of J2 in the right most area like : J1 J1 J1________| J2.
Except right side of right most J2 we can put J2 in all the gaps of J1 (gaps are shown by * star ) : * J1 * J1 * j1 * J2. So we have 4 region to put ( 4 - 1 ) 3 unit of J2 (since one unit of J2 is already put ). So we can put 3 J2 in 4 gaps in C(3 + 4 -1, 3 ) = 20 ways..

iii) Similarly we keep J3 in the right most area. Now we have 4 unit of J3 left which we want to put in 8 gaps. So the number of ways = C(11,7 ) = 330

iv) Similarly we have 3 unit of J4 left which have to put in 13 gaps. It can be done in C(15, 12 ) ways = 455 ways.

Therefore total number of Ways : 1 x 20 x 330 x 455 = 3003000.

Number (i) is unnecessary....
Hope you will understand easily...
N.B. Sorry for poor English...

[deadlineruhe]

Thanks Muba...

this post is really helpful...