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 \(20 +15 + 6 =41\).