# 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.

## 3 Answers

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.

• why did you use stars and bars in b,c and d? Why not total selection like $10^4$ in b? – idea Nov 13 '18 at 4:13
• Agreed that scoops of one kind are identical, but $10^4$ seems fine to me. Please explain... – idea Nov 13 '18 at 4:17
• @omega: $10^4$ assumes that the bowl has $4$ numbered hollows, whereas I'm considering different arrangements of the same scoops in the bowl as equal. – Christian Blatter Nov 13 '18 at 9:05
• even scoops arrangement matters...never thought that.. – idea Nov 13 '18 at 9:06
• yup, got it. Thanks for explanation... – idea Nov 13 '18 at 9:07

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.

• 2 flavours may be chosen from remaining 10 in d – idea Nov 13 '18 at 4:15
• I read d to say there are four distinct flavors, so you need to choose two of eight. I agree it is not clear. a required four distinct flavors and b did not, so there is no tradition to fall back on. If you allow extra scoops of lemon or mango, you should allow three lemon and one mango which you will not count if you choose two from ten. – Ross Millikan Nov 13 '18 at 4:28
• Yeah, so i included both cases... – idea Nov 13 '18 at 4:29

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

• It's not clear to me what you mean by "The 4 kinds are not necessarily different". If they are not different in what sense are there 4 "kinds"? – user247327 Nov 12 '18 at 18:46
• @user247327 he means that a bowl needs all 4 different flavours, not 2 scoops of 1 flavour – idea Nov 12 '18 at 18:47