Coin Distribution Problem- Combinations In how many ways can we distribute a single quarter, a single dime, a single nickel and 25 separate cents between $5$ children if
a) we have no restrictions?
b) such that the oldest kid gets either $20$ cents or $25$ cents.
a) $$\;\binom{1+5-1}{1}\times \binom{1+5-1}{1}\times \binom{1+5-1}{1}\times  \binom{25+5-1}{5} = 14,844,375$$
b) My textbook is a spanish translation of the original and the exercise is badly worded and I cannot find it in the english version, so I am confused.
Please help me out.

Assuming the children are not identical, the distribution of the 3 coins worth more than a penny can happen $5^3$ ways.  Then 25 pennies distributed among 5 children, by the "stars and bars" argument, would be ${25+5-1 \choose 4}$ so for the answer to (a) I get $125 \times \frac{29\cdot 28\cdot 27 \cdot 26}{24} = 2968875$
For part (b), first consider the case where the oldest kid gets $20$ cents.  That means he got a dime and $10$ pennies, or a dime and a nickel and $5$ pennies, or a nickel and $15$ pennies, or $20$ pennies.  I don't see any more clever way of doing this than by calculating the ways for each branching possibility.

*

*If the oldest got a dime and ten pennies:  There are four children left. So the quarter and the nickel have $4^2$ ways to go.  Then there are 15 pennies and 4 children, making ${18 \choose 3}$ ways.  So that gives $\frac{16 \cdot 18 \cdot 17 \cdot 16}{3!} = 13056$ possibilities.

The rest are calculated similarly.
