Calvin's Cream Conundrum

There are clearly 3 different 'types' of combinations: 3 scoops of the same flavour, 2 of the same flavour and 1 different, or all 3 scoops having different flavours.

For the first case (3 same flavour), there are obviously only 12 combinations.

For the second case (2 same, 1 different flavour), there are ${12 \choose 2}$ ways to choose the 2 different flavours, and for each of these cases there are 2 ways to decide which flavour gets the 'double-scoop). Therefore there are a total of $2\times {12\choose 2} = 132$ combinations.

For the third case, (3 different flavours), there are $12 \choose 3$ ways to choose 3 different flavours, which gives 220.

In total, there are $12 +132 + 220 = 364$ combinations.

for this solution, we may look into three difference condition:
1.the three scoop of ice cream is totaly difference.
2.only two scoop of ice cream are same
3.all of the three scoop of ice cream are the same

for condition no 1...
if consider the order of arrangement , the number of way of arrangement would be
n= 12 * 11 * 10
but order of arrangement is not considered in this question n no of way to arrange 3 different thing would be 3P3=6
therefore , n=(12 * 11* 10)/6=220..

for condition no 2
there will be two same ice cream...let's fix the two and there will be 11 ways to put in ..next, there are 12 different ice cream
therefore, the number of way of arrangement , n=12*11=132

for condition no 3
there would definately be 12 ways of arrangement

as a result, total of ways of arrangement
n= 220+132+12=364

The style of combinations can be summarized as: ABC for three (3) different flavors in a cup AAB for two (2) the same flavor in a cup AAA for three (3) the same flavor in a cup

Since there are 12 flavors, there is a total of ${12 \choose 3} = 220$ combinations of three (3) different flavors in a cup.

For the two (2) and three (3) the same flavor in a cup, two (2) the same flavor shall be considered as one (1) flavor. i.e, AA where this combination can be any of the 12 flavors. This follows that there are also 12 flavors that can be the third flavor to be combined with this (including also A, which make an AAA combination). Thus, for this situation, we can have $12*12 = 144$ combinations.

Therefore, there are only $220 + 144 =$ 364 combinations of 3-in-a-cup ice cream in that particular ice cream store.

Basically there can be three different ways to make a 3 in a cup: one in which all scoops are of different flavor, one in which two are of the same flavor, and one in which all flavors are different.

No. of ways of making 3 in a cup with all flavors different= No. of ways of choosing 3 different objects out of 12 different objects= 12C3=220

No. of ways of making 3 in a cup with 2 flavors of same kind= 12 * 11= 132

No. of ways of making 3 in a cup with all flavors same= 12

Adding these three values we get

Total no. of different 3 in a cup combinations= 220+ 132+ 12= 365

Let's condier 3 cases;

Case 1: 1 type of flavor

If there is only 1 type of flavor, then there can only be 12 possibilities only.

Case 2: 2 types of flavor

We need to select two flavors from 12, so ${12 \choose 2}$ . However will need to also note the number of scoops for each ice-cream respectively. Let the first flavor be A and the other be B. So we can have 2 scoops of A and 1 scoop of B OR 1 scoop of A and 2 scoops of B. Since there is two ways of scooping them, then the total possibilities for this case is; $2 \times {12 \choose 2}=132$

Case 3: 3 types of flavor

The number of scoops for each flavor can only be 1. Hence, we will only need to select three types of flavors from 12 or ${12 \choose 3}=220$ .

Therefore, the total combinations for the ice-creams would be $12+132+220=364$

$\text{Number of 1-flavour cups: }{12 \choose 1} = 12$ $\text{Number of 2-flavour cups: }2 \times {12 \choose 2} = 132$ $\text{The 2 factor is due to the difference in the choice of the double scoop}$ $\text{Number of 3-flavour cups: }{12 \choose 3} = 220$ $\text{Total = } 12+132+220 = 364$

1) All different flavors - 12C3 2) different flavors - 12C2 in this two ways so total - 2*12C2 3) Same flavor - 12C1 Total - 364

We may choose: 1 different flavors (case 1), 2 different flavors (case 2), and 3 different flavors (case 3).

Case 1: $_12C_1 = 12,$

Case 2: Let us say that the first flavor to be chosen will be 2 scoop and the later will be 1 scoop. So by $FPC$ $12 \times 11 = 132$

Case 3: Choosing 3 from 12 flavors regardless of the order is simply $_{12}C_3 = \frac{12!}{(12-3)! \cdot 3!} \Rightarrow _{12}C_3 = 220$

Adding all cases: $12 + 132 + 220 = 364$

Therefore there are $364$ combinations of 3-in-a-cup ice cream.

* NOTE: It doesn't imply that the shop owner is a liar, maybe the shop owner simply neglect to do math first. :P

classification 1- with the 3 scoops of the same flavor.................. 12 classification 2- with all the 3 scoops in different flavors . . . . . .

                                              12C3  =                                          220

classification 3- with only 2 of the three scoops of the same flavor 112,113,114,115...1112 = 11 221,223,224...2212 = 11

We consider the following cases, based on the number of distinct flavors in the cup.

Case 1. 3 distinct flavors. Since order doesn't matter, there are ${12 \choose 3} = \frac { 12 \times 11 \times 10 } {6} = 220$ such 3-in-a-cup.

Case 2. 2 distinct flavors (1 repeated twice). There are 12 choices for the repeated scoop, and 11 choices for the single scoop flavor, hence a total of $12 \times 11 = 132$ such 3-in-a-cup.

Case 3. 1 distinct flavors (repeated thrice). There are 12 chocies for the repetaed scoop.

Hence, there is a total of $220 + 132 + 12 = 364$ such 3-in-a-cup.

Note: There will be $12 \times 12 \times 12 = 1728$ only if the order of scooping matters. However, since order doesn’t matter, this Rule of Product method would double count certain types of scoops.