Ice Cream Scoops - In shops with lots of ice-cream flavors there are many different flavor combinations, even with only a 2-scoo

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Ice Cream Scoops - In shops with lots of ice-cream flavors there are many different flavor combinations, even with only a 2-scoo

p cone. With 1 ice-cream flavor there is 1 kind of 2-scoop ice cream, but with 2 flavors there are 3 possible combinations (eg vanilla/vanilla, chocolate/chocolate, and vanilla/chocolate). How many possible 2-scoop cone combinations are there with 3 flavors? How about 4 flavors? 5 flavors? Use your results to develop a pattern or formula and use it to predict the number of 2-scoop combinations with 10 flavors.

Mathematics

1 answer:

PIT_PIT [12.4K] · Answer 1 · 2026-04-03T09:09:23+03:00

Answer:

The universal formula applicable for determining the necessary combinations of n ice cream flavors.

$N_n = \frac{n(n+1)}{2}$

For 10 flavors of ice cream

$N_{10} = \frac{10(10+1)}{2}$

$N_{10} = 55$

Step-by-step explanation:

Let's denote V = Vanilla, C = Chocolate, S = Strawberry, P = Pineapple, B = Berry

For 2 flavors of ice cream:

V, C

{V/V, V/C, C/C}

3 possible combinations

For 3 flavors of ice cream:

V, C, S

{V/V, V/C, V/S, C/S, C/C, S/S}

6 possible combinations

For 4 flavors of ice cream:

V, C, S, P

{V/V, V/C, V/S, V/P, C/S, C/P, S/P, C/C, S/S, P/P}

10 possible combinations

For 5 flavors of ice cream:

V, C, S, P, B

{V/V, V/C, V/S, V/P, V/B, C/S, C/P, C/B, S/P, S/B, P/B, C/C, S/S, P/P, B/B }

15 possible combinations

Thus, we derive a sequence of

3, 6, 10, 15...

This sequence is recognized as the Triangular number series

1 + 2 = 3

1 + 2 + 3 = 6

1 + 2 + 3 + 4 = 10

1 + 2 + 3 + 4 + 5 = 15

The general formula for this sequence is

Utilizing this formula allows us to predict the number of 2-scoop combinations from 10 flavors.

$N_n = \frac{n(n+1)}{2}$

$N_{10} = \frac{10(10+1)}{2}$

$N_{10} = 55$

Consequently, there are 55 varying combinations for a 2-scoop cone when selecting from 10 flavors.