[Math] In how many ways can 10 (identical) dimes be distributed among five children

combinatoricsdiscrete mathematics

a. If there are no restrictions?

b. Each child gets at least one dime?

c. The oldest child gets at least two dimes?

For part (a), the textbook gives the answer $14 \choose 10$.

Where did the 14 come from? I thought we had 10 dimes to choose from, and 5 children to distribute them to. Why doesn't this come out to $10 \choose 5$?

Thanks!

Best Answer

The $14$ comes from the stars and bars approach, counting the number of arrangements with $5-1=4$ bars and $10$ stars. The correspondence can be illustrated by the following example: $$\overbrace{\star\star|\star\star\star\star|\star||\star\star\star}^{\text{stars and bars}} \quad \longleftrightarrow \quad \overbrace{(2,4,1,0,3)}^{\text{who gets what dimes}}.$$ There are $\binom{14}{10}$ such stars-and-bars arrangements.

Instead, $\binom{10}{5}$ is the number of ways of choosing $5$ dimes from $10$ dimes, which doesn't correspond with how to distribute them to the children.

For the second part of the question, we do the same thing after giving one dime to each child. For the third part of the question, we give the oldest student two dimes, then do the same thing again (with $8$ dimes).