Counting problem.How many different bowls can be made if: An ice-cream shop sells ten kinds of ice-cream, including mango and lemon. For a bowl, one chooses at random 4 kinds.How many different bowls can be made if:a)The 4 kinds are different
b)The 4 kinds are not necessarily different;
c)The bowl contain lemon, but no mango?;
d)The bowl contains both lemon and mango.
For point a) I thought of associating each kind of ice-cream to a number, from 1 to 10 and then I thought how many bowls can I create in such a way that each number will appear only once per bowl and the same set of numbers will not be repeated:
{1,2,3,4},{1,2,3,5}...{1,2,3,10}
{1,2,4,5},{1,2,4,6}...{1,2,4,10}
...
{1,2,8,9},{1,2,8,10}
{1,2,9,10}
and the result would be 1+2+3+4+5+6+7=28 (but I don't really like this method, and I'm not even sure if it's correct)
but for the rest I am completely clueless so I would really appreciate some help 
P.S. I'm new to counting problems so if you find any mistakes in my way of thinking please do tell me, I really want to learn how to think this kind of problems but I'm having a hard time finding solved examples.
 A: For a you just need to choose four flavors out of ten, so ${10 \choose 4}=210$  Your hand count assumed lemon and mango were required, so is the answer to d.  The easy way for d is that you have to choose two flavors of the remaining eight and ${8 \choose 2}=28$.  For c, you need to choose (how many) flavors out of (how many) to complete the bowl?.  For b, I don't know an easy way except to catalog the partitions of $4$ and figure out how many different bowls correspond to each.  Note that you should count lemon,lemon,mango,chocolate as different from mango,mango,lemon, chocolate.
A: I'm assuming that it is allowed to choose several scoops of the same kind, if not stipulated otherwise.
(a) There are ${10\choose4}$ ways.
(b) We may choose freely $4$  scoops from $10$ kinds. Therefore we have to count the number of solutions to $x_1+x_2+\ldots+x_{10}=4$ in nonnegative integers. By stars and bars this  number is ${9+4\choose4}={13\choose4}=715$.
(c) Let the first scoop be lemon. Then we may choose freely $3$ additional scoops from $9$ kinds. This can be accomplished in ${8+3\choose3}={11\choose 3}=165$ ways.
(d) Let the first two scoops be lemon and mango. Then we may choose freely $2$ additional scoops from $10$ kinds. This can be accomplished in ${9+2\choose2}={11\choose 2}=55$ ways.
A: It will be a lot easier to use combinations here.
For $a$:
Select $4$ out of $10$ in $^{10}C_4$ ways.  
For $b$:
Select first scoop in $10$ ways, second in $10$,..so on. It gives $10\cdot 10\cdot 10 \cdot10=10^4$ ways  

cases c and d are a bit ambiguous!

For $c$:
$\hspace{1.5cm}$Case $1$: A bowl has all different flavours.   A bowl has $4$ different flavours:
 Lemon is compulsory, Mango is excluded. So, you have fixed $1$ scoop, and now select rest $3$ from remaining $8$, (after excluding Lemon and Mango),
in $^8C_3$ ways.
$\hspace{1.5cm}$Case $2$: Same flavour allowed:
Here, you get to repeat Lemon scoop. So, $9^3$
For $d$:
$\hspace{1.5cm}$Case $1$: All different flavours.
$2$ are already fixed. So just select rest $2$ from remaining $8$, in $^8C_2$ ways.
$\hspace{1.5cm}$Case $2$: Same flavour allowed.
Now, you can repeat both $Lemon$ and $Mango$ 
 again. So, $10^2$

So, answers are: $$^{10}C_4,\hspace{0.5cm}10^4,\hspace{0.5cm}(^8C_3 \hspace{0.5cm}or\hspace{0.5cm}9^3)\hspace{0.5cm} and \hspace{0.5cm}(^8C_2\hspace{0.5cm}or\hspace{0.5cm}10^2)$$

