There r 100 different balls in a bag.A person takes out balls one by one and then he puts them back.Next he takes out 10 balls at a time and the rest 90 one by one.He again puts them back.He next takes out 20 balls at a time and the rest 80 one by one.He repeats the same routine until he takes 100 balls at a time.Find the total number of ways in which a person can do these operations.
Apr 20th 2011, 09:24 AM
I'm not sure I understand the question exactly, but here's my understanding: There are 100! ways to take them out one-at-a-time. There are 100 C 10 ways to choose the first ten, then of the remaining 90 there are 90! ways to choose those. Using this logic throughout, we would get 100! + (100 C 10)*90! + (100 C 20)*80! + ... + (100 C 90)*10!.
Hope this is the right way to understand the problem.
[Edit: Actually, rather than adding these we should probably multiply them since a difference in the way of doing the first selection constitutes a different overall process of selection.]
Apr 20th 2011, 09:32 AM
i also thought of the same it shld be multiplied i guess... becoz all the events are independent.... confirmed
Apr 20th 2011, 09:34 AM
also is there any way of calculating the exact figure.... exactly... Is there any such calculator that is availabel that can do it...
Apr 20th 2011, 10:04 AM
I believe it should be multiplied.
If you want to calculate this, go to wolframalpha.com and type into the text box 100!*(100choose10)*90!*(100choose20)*80!*...
Or you can try to enter it by pi notation, product((100choose10n)(100-10n)!) from n=0.. 9
I'm not sure if entering either of these will work. The first one might be too long to enter, and I'm not sure if the program will read the second one correctly. But if you play with it I think you can figure something out.