there are 20 books and 10 boys. each boy must at least get 1 book. in how many ways can i distribute these books?

Hi Aritra,

Firstly each boy must get atleast one book.

So 10 books are given, one each to one boy.

Now we are left with 10 books, to be distributed to 10 boys with no condition being imposed on the distribution.

Let Xi be the number of books the i^th boy gets.

So X1+X2+X3+......X10 = 10 ; where 0≤Xi≤10.

The number of solutions to this equation is ¹⁹C₁₀. (which is also same as ¹⁹C₉).

The above solution is on the assumption that all the 20 books are identical.

All the best.

Regards,

Ashwin (IIT Madras).

ⁿC_r = n!/r!(n-r)!

     = 20!/10!(10)!

first we select 10 out of 20 by²⁰C₁₀ then arrang it  as ^10P₁ 

SO  the possible way is = ²⁰C₁₀* ¹⁰P₁

                                 =(20!/10!*10!)* 10!/9! WAYS