Myujth

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.

I'm assuming that it is allowed to choose several scoops of the same kind, if not stipulated otherwise.

(a) There are $10choose4$ 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 $9+4choose4=13choose4=715$.

(c) Let the first scoop be lemon. Then we may choose freely $3$ additional scoops from $9$ kinds. This can be accomplished in $8+3choose3=11choose 3=165$ 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 $9+2choose2=11choose 2=55$ ways.

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.

It will be a lot easier to use combinations here.

For $a$:

Select $4$ out of $10$ in $^10C_4$ ways.

For $b$:

Select first scoop in $10$ ways, second in $10$,..so on. It gives $10cdot 10cdot 10 cdot10=10^4$ ways

cases c and d are a bit ambiguous!

For $c$:
$hspace1.5cm$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.
$hspace1.5cm$Case $2$: Same flavour allowed:

Here, you get to repeat Lemon scoop. So, $9^3$

For $d$:
$hspace1.5cm$Case $1$: All different flavours.
$2$ are already fixed. So just select rest $2$ from remaining $8$, in $^8C_2$ ways.
$hspace1.5cm$Case $2$: Same flavour allowed.

Now, you can repeat both $Lemon$ and $Mango$
again. So, $10^2$

So, answers are: $$^10C_4,hspace0.5cm10^4,hspace0.5cm(^8C_3 hspace0.5cmorhspace0.5cm9^3)hspace0.5cm and hspace0.5cm(^8C_2hspace0.5cmorhspace0.5cm10^2)$$