COMBINA

Returns the number of combinations of a subset of items.

Syntax:

COMBINA(n, k)

n is the number of items in the set. k is the number of items to choose from the set. COMBINA returns the number of unique ways to choose these items, where the order of choosing is irrelevant, and repetition of items is allowed. For example if there are 3 items A, B and C in a set, you can choose 2 items in 6 different ways, namely AA, AB, AC, BB, BC and CC; you can choose 3 items in 10 different ways, namely AAA, AAB, AAC, ABB, ABC, ACC, BBB, BBC, BCC, CCC. COMBINA implements the formula:(n+k-1)!/(k!(n-1)!)

Example:

COMBINA(3,2)

returns 6.

COMBINA(3,3)

returns 10.

Application:

Ice Cream Scoops

Imagine a customer at an ice cream shop wants to create a two-scoop cone. The shop has four available flavors:

Vanilla
Chocolate
Strawberry
Mint

Since the customer can choose the same flavor twice (e.g., two scoops of vanilla), this is a combination with repetition problem.

Here's a table illustrating the different combinations of two scoops:

	Combination	Scoop 1	Scoop 2
	A	B	C
1	1	Vanilla	Vanilla
2	2	Vanilla	Chocolate
3	3	Vanilla	Strawberry
4	4	Vanilla	Mint
5	5	Chocolate	Chocolate
6	6	Chocolate	Strawberry
7	7	Chocolate	Mint
8	8	Strawberry	Strawberry
9	9	Strawberry	Mint
10	10	Mint	Mint

There are 10 possible combinations.

The COMBINA function would be written as:

COMBINA(n, k)

Where:

n is the number of flavors (4)
k is the number of scoops (2)

Using the formula, we get:

COMBINA(4, 2)

Result:

10

This will return the value 10, which matches the number of combinations in the table.

The mathematical formula for combinations with repetition is:

$\frac{( n + k - 1 )!}{k ! ( n - 1 )!} = (k n + k - 1)$

Substituting our values:

$\frac{( 4 + 2 - 1 )!}{2 ! ( 4 - 1 )!} = \frac{5 !}{2 ! 3 !} = \frac{120}{( 2 ) ( 6 )} = \frac{120}{12} = 10$