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 flavors is unimportant?
$$\frac{31!}{3!28!}$$
b)each flavor must be different and the order of flavors is important?
$$\frac{31!}{28!}$$
c)Flavors need not be different and the order of flavors is unimportant?(This is a non-trivial question)
$$\frac{31!}{3!30!}$$
d) Flavors need not be different and the order of flavors is important?
$$31^3$$
Could you check it for me please?
Best Answer
Yes. $^{31}C_3$ or $\binom{31}{3}$ counts the ways to select 3 unique items from 31.
Likewise, $^{31}P_3$ or $\binom{31}{3}3!$ is the ways to select 3 from 31 and arrange them.
Indeed! The "stars and bars" method counts $\binom{31+3-1}{31-1}$ ways to put $3$ identical items into $31$ distinct boxes - or in this case take $3$ scoops from $31$ tubs.
Alternatively you might have counted the ways to select: three identical scoops, or a pair and a single, or three different scoops. $31+31\cdot 30+\binom{31}{3}$
Yes, $31^3$ counts the ways to make 3 independent choices with $31$ options each.