1.6.4 problem 59

(a)

$\left(\frac{15+9}{9}\right)$

(b)

$\left(\frac{5+9}{9}\right)$

(c)

Each of 15 bars can be given to any of 10 children, so by orderd sampling with replcement formula we have $10^{15}$ combinations

(d)

To count amount of suitable combinations, we can subtract amount of combination, where at least one child doesn’t get any bars (is example of inclusion-exclusion usage case) from total amount of combinations.
$10^{15}-{\sum }_{i=1}^9{\left(-1\right)}^{i+1}\left(\frac{10}{i}\right){\left(10-i\right)}^{15}$