In how many ways can $$10$$ different balls be divided between two boys, one receiving two and the other eight balls
A
$$45$$B
$$75$$C
$$90$$D
None of these
No. of distribution$$= \cfrac{10!}{2!\times 8!}=45$$
When $$B$$ gets $$2$$ and $$A$$ gets $$8$$ balls,
No. of distribution$$= \cfrac{10!}{2!\times 8!}=45$$
$$\therefore \text{Total no. of ways}=45+45=90$$.
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We don’t have your requested question, but here is a suggested video that might help.
10. How many ways are there to place 10 indistinguishable balls into eight distinguishable bins?
No of distribution = 10! /(2! * 8!) = 45 (When A get 2 and B gets 8 balls.) No of distribution = 10! /(2! * 8!) = 45 (When B get 2 and A gets 8 balls.)
So, Total Number of Ways = 45 + 45 = 90.