Buy 25 pieces of jelly of 3 flavours with at least 5 of 1 flavour. How many ways to buy the flavours?

Question

Terra, Ventus and Aqua are very good friends, almost like siblings. To celebrate Terra’s job promotion, Ventus and Aqua decided to buy Terra’s favorite dessert, fruit-flavored jelly, particularly cherry-flavored jelly. The two decided to buy a total of 25 pieces of jelly of varying flavors:

• cherry,
• blueberry and
• kiwi.

When they arrived at the jelly market, they find that there are only 9 pieces of jelly left for each of the three flavors that they want to buy.

Since it’s also Terra’s birthday, they would buy at least 5 pieces of cherry-flavored jelly.

Now, in how many ways can they pick the flavors for the remaining pieces of jelly?

Answer 1

Let $X_i, i = 1,2,3$ be the number of pieces of jelly from cherry, blueberry and kiwi respectively. They have decided that they buy 5 pieces of cherry for the birthday girl. They also decided that they will buy 25 of which 5 is cherry and the rest 20 from the remanining flavors including cherry. There are only 9 available in each flavor for 2 and 3 and only 4 available for 1 ( as she would scoop 5 already).

$x_i \le 9, i = 2,3$ and $x_1\le4$

Let $y_i = 9-x_i, i = 2,3$ and $y_1 = 4-x_1$ with $y_i's\ge0$

The classic problem is $x_1+x_2+x_3 = 20$

or $4-y_1+9-y_2+9-y_3 = 20$

or $y_1+y_2+y_3 = 2$

The number of solutions for this would be the number of ways they could buy the 20 jellies which is ${(2+3-1)\choose(3-1)} = 6$

Answer 2

i might be misinterpreting but

• case 1: 5 cherry.

under case 1 you could have

0 blueberry 20 kiwi, 1 b 19 k, ..., 19 k 1 b, 20k 0b

for a subtotal of 21 subcases.

• for case 2: 6 cherry

you have

0 blueberry 19 kiwi, 1b 18 k, ..., 19 b 0 k

for a subtotal of 20 cases.

• case 3 has 19 subcases.

• case 4: 18

• case 5: 17

soooo $21+20+19+18+17=95$

?

Or use piecewise/indicator/characteristic functions to say 21, 20, 19, 18 or 17 depending on how many cherry pieces bought.

Answer 3

Hint: $7+9+9$ and $8+8+9$ are the only ways to partition $25$ into three unlabelled heaps such that no heap is more than $9$.

So how many ways are there to select heaps of cherry, blueberry, and kiwi flavours such that no heap is more than $9$?

Answer 4

At the jelly store, they would have already taken 5 cherry flavored jellies for Tara. This leaves 4 cherry, 9 blue berry and 9 kiwi left. They still have to choose 20 more to get to 25.

If we take $c$ to be the number of cherry flavored jellies, $b$ the number of blue berry flavored jellies and $k$ to be the number of kiwi flavored jellies they choose, we have:

$$ c+b+k = 20 $$

but they can't just choose any number like 18 kiwi since the store has a limited stock.

Our constraints are:

$$ c≤4 $$

$$ b≤9 $$

$$ k≤9 $$

If we call new variables as $C$, $B$ and $W$ where $C = 4-c$, $B =9-b$ and $K = 9-k$, (as these new variables will always be non-negative) our equation will now be:

$$C +B + K = 2$$

That has the solutions $(C,B,K)$ = $(2,0,0)$,$(0,2,0)$,$(0,0,2)$,$(0,1,1)$,$(1,0,1)$ and $(1,1,0)$.

So, 6 ways.