Four children

The following are the possible combinations: (1,1,3,3), (1,2,2,3), (2,2,2,2)

Now, (1,1,3,3) can be arranged in 4!/2!2! = 6 ways

(1,2,2,3) can be arranged in 4!/2! = 12 ways

(2,2,2,2) can be arranged only in 1 way.

So, total no of ways = 6 + 12 + 1 = 19

We can see that each child can have either 1,2 or 3 rupees.

Let us take the series

$(x+{ x }^{ 2 }+{ x }^{ 3 })$

The power of the term signifies the no. of rupees a child has.

For 4 children, we take the product of 4 such series:

$(x+{ x }^{ 2 }+{ x }^{ 3 })(x+{ x }^{ 2 }+{ x }^{ 3 })(x+{ x }^{ 2 }+{ x }^{ 3 })(x+{ x }^{ 2 }+{ x }^{ 3 })$

which is basically

${ (x+{ x }^{ 2 }+{ x }^{ 3 }) }^{ 4 }$

Now we must choose one term from each expansion such that the product of our chosen terms must contain ${ x }^{ 8 }$ (Since there are a total of 8 rupees.)

Now expanding the earlier expression we have to find the coefficient of ${ x }^{ 8 }$ which turns out to be 19.