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 different

So 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 same

You have to choose 1 topping = \(6\) (\(C^6_{1}\))

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

Hi..

if we have no repetitions it is same as choosing 3 out of 6 so 6C3=\(\frac{6!}{3!3!}=\frac{6*5*4}{3*2}=20\)

Now when repetitions are allowed, following will get added..
1) Any one repeated twice..
so choose two out of 6 = 6C2 = \(\frac{6!}{4!2!}=15\)
But each choice will have two ways as one of them can be taken twice....
for example if two choosen is A and B... this will have two ways AAB or ABB
thus total = 15*2=30
2) any one repeated thrice so 6 ways

total = \(20+30+6=56\)

In response to: msawicka

Basically, combinations with replacement or possible repeats (choices are put back in the pool of choices) can be solved by replacing n with n+k-1, then solving the problem like any typical nCk problem.

For example, in this problem we have 6 choose 3. If repetition or repeats are allowed, replace n with n+k-1

original n = 6

new n = n+k-1 = 6+3-1 = 8

\(\frac{n!}{(n-k)!k!}\)

\(\frac{8!}{5!3!}\)

\(\frac{8x7x6x5!}{5!3!}\) {5!'s cancel out}

\(\frac{8x7x6}{3!}\)

\(\frac{8x7x6}{3x2x1}\)

\(\frac{8x7x6}{6}\) {6's cancel out}

8x7 = 56

Here is a video which explains how to solve combination problems with repetition or replacement: https://www.youtube.com/watch?v=ZcSSI6V ... =17&t=24s#

Having said all this, I don't believe combination with repetition or repeats are on the GRE.

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