The sundae hut offers $6$ different toppings for their sundaes. If you can get a sundae with any number of toppings you want (including no toppings), how many different sundaes can you get?

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Details and assumptions
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You cannot get the same topping twice.

The order of toppings doesn't matter.

The answer is 64.

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I converted this comment into a solution, as it shows how choosing the correct perspective in counting a combinatorics question can greatly simply the work.

for me any of those solutions are right, you can have diferent orders of sunday, like.

123456 123465 123645 126345 162345 612345 12345 12354 12534 15234 51234 1234 1243 1423 4123 132 312 21

and you have that for each one like

612345 612354 ... so the solution to this problem is wrong

juan tenorio
- 7 years, 7 months ago

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only if they first said that the order dont matter, the topping are the same your solution could be right

juan tenorio
- 7 years, 7 months ago

Okay .

For the following problem ,

we begin by doing
**
casework
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on the
*
number of toppings
*
we want out of
$6$
.

$\begin{array}{c|c|c} & Toppings & Combinations = ^nC_r = \dfrac{n!}{r!(n-r)!}\\ \hline Case-1 & 0 & ^6C_0 = 1 \\ Case-2 & 1 & ^6C_1 = 6 \\ Case-3 & 2 & ^6C_2 = 15\\ Case-4 & 3 & ^6C_3 = 20\\ Case-5 & 4 & ^6C_4 = 15\\ Case-6 & 5 & ^6C_5 = 6\\ Case-7 & 6 & ^6C_6 = 1 \end{array}$

Thus ,

**
Total number of cases
**
are -

$^6C_0 + ^6C_1 + ^6C_2 + ^6C_3 + ^6C_4 + ^6C_5 + ^6C_6$ $= 1 + 6 + 15 + 20 + 15 + 6 + 1 = \boxed{64}$

Thus , we can get $\boxed{64}$ sundaes ! (Yumm..)

Cheers!

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Is there a simpler way of approaching the problem without using 7 cases?

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Yes. Knowing the "line" Pascal's Triangle pattern, we can generalize the problem to $\sum_{n=0} ^{x} \binom{x}{n} = 2^n$ .

I consider very saddening the fact that those who posted a solution didn't know that, even if it's a L1/L2 problem. :(

Guilherme Dela Corte
- 7 years, 7 months ago

I guess this is the intended way, each topping can either be included or excluded on the sundae, hence by the rule of product there are $2^6$ ways.

Jan J.
- 7 years, 7 months ago

I realized that everyone had used casework only after seeing the solutions, so I posted a solution of sorts in the form of a comment in the solution Aditya wrote above. I am quite certain that it works for any value of n(=6 for this question).

Rohan Rao
- 7 years, 7 months ago

Please don't worry about the scroll . There's nothing hidden (!)

Priyansh Sangule
- 7 years, 7 months ago

I don't get the scroll, but thanks anyway.

Pola Forest
- 7 years, 7 months ago

excellent description friend

Absec Panda
- 7 years, 7 months ago

Ways to choose sundae with no toppings = 1

Ways to choose sundae with 1 topping = 6C1 = 6

Ways to choose sundae with 2 toppings = 6C2 = 15

Ways to choose sundae with 3 toppings = 6C3 = 20

Ways to choose sundae with 4 toppings = 6C4 = 15

Ways to choose sundae with 5 toppings = 6C5 = 6

Ways to choose sundae with 6 toppings = 6C6 = 1

Hence, total number of sundaes possible = 1+1+6+15+20+15+6+1 = 64.

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A slightly swifter solution:

Consider that for each toping, you can either have it or not have it. Thus there are $2$ possibilities for each of $6$ toppings. Hence there are $2*2*2*2*2*2=2^6=64$ possible sundaes.

William Mitchell
- 7 years, 7 months ago

Thanks! Did you work all that out in your head, though?

Pola Forest
- 7 years, 7 months ago

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Yes , we all had to. :)

Priyansh Sangule
- 7 years, 7 months ago

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There IS a simpler way of doing that. Please, check my comment under Calvin's comment on your solution.

Guilherme Dela Corte
- 7 years, 7 months ago

Aww. Okay ,thank you so much :)

Jacgil Antoni Oculam
- 7 years, 7 months ago

easy,it's simple combination application

6C0 +6C1 +6C2 + 6C3 + 6C4 +6C5 +6C6=64

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In Charak Sanhita, Charak mentioned that if we have 6 things we can take them in 63 ways not including the nothing case. So 63+1 = 64

Jitesh Mittal
- 7 years, 1 month ago

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the formula for the amount of combinations with no repeats and where each answer is distinct is /frac{n!} {(n-r)!(r)!} where n is the number to choose from and r is the number chosen n is always 6, so we plug it in for each possible value of r /frac{6!}{(6-6)!(6)!}=1

/frac{6!}{(6-5)!(5)!}=6

/frac{6!}{(6-4)!(4)!}=15

/frac{6!}{(6-3)!(3)!}=20

/frac{6!}{(6-2)!(2)!}=15

/frac{6!}{(6-1)!(1)!}=6

and the option of no toppings=1

add them to gether and we get 1+6+15+20+15+6+11=
**
64
**

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"Is there a simpler way of approaching the problem without using 7 cases?" - Calvin L.

Guilherme Dela Corte
- 7 years, 7 months ago

In order to get a sundae, you can have anywhere between 0 and 6 toppings. The number of ways of selecting the following combinations: - 0 toppings = 6C0 = 1 - 1 topping = 6C1 = 6 - 2 toppings = 6C2 = 15 - 3 toppings = 6C3 = 20 - 4 toppings = 6C4 = 6C2 = 15 - 5 toppings = 6C5 = 6C1 = 6 - 6 toppings = 6C6 = 6C0 = 1

Total on adding all the possible combinations = \boxed{64}

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Can be more systematic using Latex

Priyansh Sangule
- 7 years, 7 months ago

Total ways to select all that is summation(6Cr) for r belonging to 0 to 6 which is just 2 raised to 6. If the number were, say, 25, surely all of us wouldn't be counting cases! :) Actually, (1+x)^6 is the generating function since we can either have a topping or not have the topping and the total number is 6. We need the sum of the coefficients, so we substitute x as 1 to get 2^6 or 64.

Rohan Rao
- 7 years, 7 months ago

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Good, those are other ways of approaching the problem, where we don't necessarily have to evaluate all of the cases to know that the sum is $2^6$ . There is still a much simpler way (generating functions would be an overkill for this problem).

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Hint:
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Rule of product.

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Consider that for each toping, you can either have it or not have it. Thus there are $2$ possibilities for each of $6$ toppings. Hence there are $2*2*2*2*2*2=2^6=64$ possible sundaes.