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.
Best Answer
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.