In how many ways can 15 identical blankets be distributed among six beggars such that everyone gets at least one blanket and tow particular beggars get equal blankets and another three particular beggars get equal blankets.

Explanation

The number of ways of distributing blankets is equal to the number of solutions of the equations `3x+2y+z=15`.
where `xge1,yge1,zge1` which is equal to coefficient of `alpha^(15)` in
`(alpha^(3)+alpha^(6)+alpha^(9)+alpha^(12)+alpha^(15)+ . ..)xx(alpha^(2)+alpha^(4)+alpha^(6)+alpha^(8)+alpha^(10)+alpha^(12)+alpha^(14)+ . .)`
`xx(alpha+alpha^(2)+alpha^(3)+ . .+alpha^(15)+ . . .)`
=Coefficient of `alpha^(9)` in `(1+alpha^(3)+alpha^(6)+alpha^(9))xx(1+alpha^(2)+alpha^(4)+alpha^(6)+alpha^(8))`
`xx(1+alpha+alpha^(2)+alpha^(3)+alpha^(4)+alpha^(5)+alpha^(6)+alpha^(7)+alpha^(8)+alpha^(9))` [negletcing higher powers]
=Coefficient of `alpha^(9)` in `(1+alpha^(2)+alpha^(4)+alpha^(6)+alpha^(8)+alpha^(3)+alpha^(5)+alpha^(7)+alpha^(9)+alpha^(6)+alpha^(8)+alpha^(9))xx(1+alpha+alpha^(2)+alpha^(3)+alpha^(4)+alpha^(5)+alpha^(6)+alpha^(7)+alpha^(8)+alpha^(9))` [neglecting higher powers]
`=1+1+1+1+1+1+1+1+1+1+1+1=12`
Case II If the inequation
`x_(1)+x_(2)+x_(3)+ . .+x_(m)len` . .. (i)
[when the required sum is not fixed]
In this case, we introduce a dummy variable `x_(m+1)`. so that,
`x_(1)+x_(2)+x_(3)+ . ..+x_(m)+x_(m+1)=n`
`x_(m+1)ge0`
Here, the number of sols of eqs. (i) and (ii) will be same.