How many ways can 10 balls be distributed to two boys one receiving two and the other eight balls?

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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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.