msawicka wrote:

How would this be different if you were allowed to repeat toppings?

To the same question without repeating toppings:

Topping 1: 6 choices

Topping 2: 5 choices (6-1 toppings as one topping is already selected)

Topping 3: 4 choices (6-2 toppings as two toppings are already selected)

Total= \(6 \times 5 \times 4 = 120\)

Now say the three toppings are jelly, jam and nuts.

So 120 will contain {jelly, jam, nut},{nut,jelly, jam} ..... and 4 other combinations. Numbers of ways of arranging 3 things \(3!=6\)

So the correct number of options for pancakes =\(\frac{120}{6}=20\).

For repeating allowed you need to you need to add two more cases

2 toppings are same one differentSo you have to choose 2 toppings and the order does not matter= \(\frac{6 \times 5}{2!}=15\) (\(C^6_{2}\))

Here you can repeat the first topping 2 times or the second topping twice so

Total choices= \(15 \times 2\)

All 3 toppings sameYou have to choose 1 topping = \(6\) (\(C^6_{1}\))

Add them all up \(20 +15 + 6 =41\).

_________________

Sandy

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