1)In how many ways can 12 balls be divided among 3 people such each can get a max of 6 balls .2) a +b +c+ d = 90 , a,b,c,d Let me know how to take care of this max condition ?
1) A...B...C
3...3...6 -> 12c3*9c3*3!/2!
3...4...5 -> 12c3*9c4*3!
4...4...4 -> 12c4*8c4
2...5...5 -> 12c2*10c5*3!/2!
2...4...6 -> 12c2*10c4*3!
1...6...5 -> 12c1*11c6*3!
hope i haven't missed any case...
2) (45-a) + (45-b) + (45-c) + (45-d) = 90
aese karke karte hai iss wale ko....koi aur method ho kisi ke pass toh batana...
PS - I am in office that is why not solving it completely...srry for that..
1) A...B...C3...3...6 -> 12c3*9c3*3!/2!3...4...5 -> 12c3*9c4*3!4...4...4 -> 12c4*8c42...5...5 -> 12c2*10c5*3!/2!2...4...6 -> 12c2*10c4*3!1...6...5 -> 12c1*11c6*3!hope i haven't missed any case...2) (45-a) + (45-b) + (45-c) + (45-d) = 90aese karke karte hai iss wale ko....koi aur method ho kisi ke pass toh batana... PS - I am in office that is why not solving it completely...srry for that..
Even I'm not sure about both of these...
For 2nd one:
according to your solution, a+b+c+d=90 which gets us back to original equation..And then ans will be 93C3...
But if we think logically it should be less than that no???due to the max. condition some cases wont appear in ans...
1)In how many ways can 12 balls be divided among 3 people such each can get a max of 6 balls .2) a +b +c+ d = 90 , a,b,c,d Let me know how to take care of this max condition ?