Combinations and Permutations in ice cream cones… what is the difference?

Question

The first two I am certain I have correct however c and d... I am struggling to understand the difference. I have done research on this site and have seen similar questions with explanations but I just want to double check my thinking.

A certain store sells 31 different flavors of ice cream. How many different 3-scoop cones are possible if:

a. each flavor must be different and the order of the flavors is unimportant? $31 nCr 3 = 4495$

b. each flavor must be different and the order of the flavors is important? $31 nPr 3 = 26970$

c. Flavors need not be different and the order of the flavors is unimportant? (This is a nontrivial question.) I think I need to use this formula $$\binom{n+m-1}{m}$$

where $n$ in the total numbers in the set, and m is how many numbers you want to choose.

So, $$\binom{31+3-1}{3}=\binom{32}{3}=\frac{32\times 31\times 30}{1\times 2\times 3}=4960$$ Is this correct?

d. Flavors need not be different and the order of the flavors is important? I am thinking this is $31 \cdot 31 \cdot 31 = 29791$

Answer 1

I think your (c) is wrong. You can count it by considering $3$ different cases: 1) $3$ flavors $X_3$; 2) $2$ flavors $X_2$ and 3) $1$ flavor $X_1$.

• The case of $3$ flavors is easy: $$ X_3 = {31 \choose 3} $$

• The case of $2$ flavors. You first choose $2$ flavors, then choose one of them to be the $2$ cones and another as the remaining cone. $$ X_2 = {31 \choose 2}\cdot 2 $$

• The case of $1$ flavor. $$ X_1 = {31 \choose 1} $$

Answer 2

part d) is correct part c) should be:

all different: $\binom{31}3$

two the same: $31\times30$

all the same 31

total $5456$

Answer 3

For c, you may use star-and-bar model, that is

$$\Large\star \star \star \normalsize \underbrace{||...||}_{30},$$

with $3$ stars and $30$ bars. The stars represent the $3$ scoops and the bars form $31$ slots, which represent the $31$ flavors. As the flavor can be the same and the order of the scoops is unimportant, there are

$${31+3-1 \choose 3}={33 \choose 3}$$ ways, which is the same as the sum of the $3$ cases which Echo presents. In addition, there seems to be a minor math error in your answer for c.

Answer 4

(d) is correct.

For (c), you had the correct idea and formula, but you made an arithmetic error. $31 + 3 - 1$ is $33$, not $32$. Using $\binom{33}{3}$ would give you $5456$, which is the correct answer.