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

290 Points
9 years ago

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 ith 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,

basit ali
41 Points
9 years ago

nCr = n!/r!(n-r)!

= 20!/10!(10)!

basit ali
41 Points
9 years ago

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

SO  the possible way is = 20C10* 10P1

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