How many different ways are there to choose 10 donuts from 20 varieties if…
a. (5 pt.) there are no restrictions?
b. (5 pt.) all the donuts chosen are of different varieties?
c. (5 pt.) at least 6 glazed donuts are chosen?
d. (5 pt.) at most 4 chocolate donuts are chosen?
My attempt
a)$\binom{20}{10}$
b)$\binom{29}{20}$ = $\binom{29}{9}$
c)$\binom{25}{6}$
d)$\binom{19}{0}$ +$\binom{20}{1}$ +$\binom{21}{2}$ +$\binom{22}{3}$ +$\binom{23}{4}$
Please verify my answers and also please provide any alternate solution for d.. Thanks
The solution to (a) is $\binom{29}{19}$. Imagine 10 donuts and 19 partitions between the 20 varieties. You need to choose which are partitions and which are donuts.
The solution to (b) is $\binom{20}{10}$. You just need to pick which 10 types from the 20.
The solution to (c) is $\binom{23}{19} = \binom{23}{4}$. You just have to pick flavors for the other four doughnuts. Use the same solution as (a).
(d) is probably easiest as $\binom{29}{19} - \binom{24}{19}$, since it's the answer to (a) minus the cases with at least 5 chocolate donuts.