In how many ways can $10$ identical presents be distributed among $6$ children so that each child gets at least one present.
Let $1$ present be given to each child. This can be done in $1$ way ($\because \;$ all presents are identical).
Then we are left with $4$ identical presents to be distributed among $6$ children.
This can be done as follows:
- Give all $4$ presents to one child.
This can be done in $\;$ $C^6_1 = \dfrac{6!}{1! \times 5!} = 6$ ways
OR - Give $3$ presents to one child and $1$ present to another.
This can be done in $\;$ $C^6_1 \times C^5_1 = 6 \times 5 = 30$ ways
OR - Give $2$ presents to one child and $2$ to another child; i.e. select $2$ children from $6$ children.
This can be done in $\;$ $C^6_2 = \dfrac{6!}{2! \; 4!} = \dfrac{6 \times 5 \times 4!}{2 \times 4!} = 15$ ways
OR - Give $2$ presents to one child and $1$ present each to two children.
This can be done in $\;$ $C^6_1 \times C^5_2 = 6 \times \dfrac{5!}{2! \times 3!} = 6 \times \dfrac{5 \times 4 \times 3!}{2 \times 3!} = 60$ ways
OR - Give $1$ present to one child each; i.e. select $4$ children from $6$ children.
This can be done in $\;$ $C^6_4 = \dfrac{6!}{4! 2!} = \dfrac{6 \times 5 \times 4!}{4! \times 2} = 15$ ways
$= 1 \times \left(6 + 30 + 15 + 60 + 15\right) = 126$ ways