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.
 A: 
In how many ways can we distribute a single quarter, a single dime, a single nickel, and $25$ pennies between five children if we have no restrictions?

There are five ways to distribute the quarter, five ways to distribute the dime, and five ways to distribute the nickel.  Since there are no restrictions on the distribution of the pennies, the number of ways they can be distributed is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25 \tag{1}$$
in the nonnegative integers, where $x_i$ represents the number of pennies received by the $i$th oldest child.
A particular solution of equation 1 corresponds to the placement of $5 - 1 = 4$ addition signs in a row of $25$ ones.  For instance,
$$1 1 1 1 1 1 1 + 1 1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 7$, $x_2 = 6$, $x_3 = 5$, $x_4 = 4$, $x_5 = 3$.  The number of such solution corresponds the number of ways we can select which $4$ of the $29$ positions required for $25$ ones and four addition signs will be filled with addition signs, which is
$$\binom{25 + 5 - 1}{5 - 1} = \binom{29}{4}$$
Hence, you should have obtained the answer
$$5^3\binom{29}{4}$$
When choices are made independently, you should be multiplying rather than adding.  Addition is used when choices are mutually exclusive.

In how many ways can we distribute a single quarter, a single dime, a single nickel, and $25$ pennies between five children if the oldest kids gets either $20$ cents or $25$ cents?

Consider cases, depending on how many cents the oldest child receives and how that amount is distributed.  Once you have chosen which coins the oldest child has received, distribute the remaining coins to the remaining children.
For instance, if the oldest child receives $25$ cents as by receiving the dime, the nickel, and ten pennies, distribute the quarter and the remaining $15$ pennies to the other four children.
A: 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.
