2 exceptional things in one day:-
KKR winning it for shahrukh and
Facebook(our server never sleeps) going blank for maintanence...

what a day in cricket and world history!!! by d way 3 cheers for kkr...

WE ROCK!!!

the logic for 7 identical things to 5 people-
1 st we have to find out all d cases-
eg(7,0,0,0,0)-5 ways
(6,1,0,0,0)- 5c2*2 ways
(5,2,0,0,0)-5c2*2ways
(5,1,1,0,0)-5c3*(3!/2) ways
(4,3,0,0,0)-5c2*2 ways
(4,2,1,0,0)-5c3*3! ways
and there are a few more cases-
it ends with (2,2,1,1,1)- 5!/(2!*3!) ways.

the sum of all the ways gives the soln.
easier way is a+b+c+d+e=7 11c4

hence 8*13C3*4C3*4C3

seems to be correct.
i missed out on the negative values.. btw then jst check out my way once.. i guess thr is some error in it apart from the negative value thing. @ sam

___/\___ chillfactor sir

m extremely sorry for those repeated posts..
my net wasn't working properly....

sry again!!!

factorise 15360 to get 2^10*3*5
now write all of them serially 2 2 2 2 2 2 2 2 2 2 3 5.
we have 4 cases now:-
1> no nos is 1: we select 3 partition out of 11 in 11c3 ways.
2> 1 nos is 1: select that in 4c1 ways and then select 2 out of 11 partitions.
therefore total= 4* 11c2
3> 2 nos are 1: select 2 out of 4 in 4c2 ways and then select 1 out of 11 partition apace availablein 11c1 ways.
therefore total - 4c2* 11
4> 3nos are 1: selection in 4c3 ways and the last nos has 2^10*3*5.
that is total - 4 ways

therefore sum of all d ways= 455

factorise 15360 to get 2^10*3*5
now write all of them serially 2 2 2 2 2 2 2 2 2 2 3 5.
we have 4 cases now:-
1> no nos is 1: we select 3 partition out of 11 in 11c3 ways.
2> 1 nos is 1: select that in 4c1 ways and then select 2 out of 11 partitions.
therefore total= 4* 11c2
3> 2 nos are 1: select 2 out of 4 in 4c2 ways and then select 1 out of 11 partition apace availablein 11c1 ways.
therefore total - 4c2* 11
4> 3nos are 1: selection in 4c3 ways and the last nos has 2^10*3*5.
that is total - 4 ways

therefore sum of all d ways= 455

factorise 15360 to get 2^10*3*5
now write all of them serially 2 2 2 2 2 2 2 2 2 2 3 5.
we have 4 cases now:-
1> no nos is 1: we select 3 partition out of 11 in 11c3 ways.
2> 1 nos is 1: select that in 4c1 ways and then select 2 out of 11 partitions.
therefore total= 4* 11c2
3> 2 nos are 1: select 2 out of 4 in 4c2 ways and then select 1 out of 11 partition apace availablein 11c1 ways.
therefore total - 4c2* 11
4> 3nos are 1: selection in 4c3 ways and the last nos has 2^10*3*5.
that is total - 4 ways

therefore sum of all d ways= 455

factorise 15360 to get 2^10*3*5
now write all of them serially 2 2 2 2 2 2 2 2 2 2 3 5.
we have 4 cases now:-
1> no nos is 1: we select 3 partition out of 11 in 11c3 ways.
2> 1 nos is 1: select that in 4c1 ways and then select 2 out of 11 partitions.
therefore total= 4* 11c2
3> 2 nos are 1: select 2 out of 4 in 4c2 ways and then select 1 out of 11 partition apace availablein 11c1 ways.
therefore total - 4c2* 11
4> 3nos are 1: selection in 4c3 ways and the last nos has 2^10*3*5.
that is total - 4 ways

therefore sum of all d ways= 455

is the soln to the pblm a*b*c*d=15360 :- 455?

is the soln to the rem pblm 20?