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Combination Problem

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by ket » Mon May 18, 2009 8:53 am
Luis can select one or more of the following 3 toppings for his ice cream: nuts, whipped cream, cherries. If he selects one or more, how many different combinations of toppings are possible (assume that the order of the toppings does not matter).

How can you solve this one?

The answers are:

A.1 B.3 C.5 D.7 E.9

I haven't got OA.
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by anksgupta » Mon May 18, 2009 9:00 am
got ans as D.

Total 3cases : selecting exactly 1 toping, selecting exactly 2 topings and exactly 3 topings.

Since order doesn't matters,

Case 1: 3 ways...
Case 2: 3 ways... 3C2
Case 3: 1 way 3C3

total 7
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by ket » Mon May 18, 2009 9:07 am
thanks, Now it seems pretty obvious :D
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Re: Combination Problem

by dtweah » Mon May 18, 2009 9:08 am
ket wrote:Luis can select one or more of the following 3 toppings for his ice cream: nuts, whipped cream, cherries. If he selects one or more, how many different combinations of toppings are possible (assume that the order of the toppings does not matter).

How can you solve this one?

The answers are:

A.1 B.3 C.5 D.7 E.9

I haven't got OA.
Select 1: 3C1 =3. He has 3 ways of choosing 1: Can choose, N or W, or C

Select 2: 3C2 =3 ways: can choose NW, NC or WC ( Since order not important, CN=NC for example)

Select 3. 3C3 = 1. Has only one way NWC

So 7 total Ways.

Choose D.
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by m&m » Mon May 18, 2009 9:12 am
3 toppings
each can be chosen or not chosen so 2 decisions

2 decisions * 2 decisions * 2 decisions
= 2^3 = 8

but we want to exclude the case where all 3 toppings are not chosen (because we know at least 1 topping IS chosen) so 8-1 = 7

hope that helps
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by ket » Mon May 18, 2009 9:29 am
m&m wrote:3 toppings
each can be chosen or not chosen so 2 decisions

2 decisions * 2 decisions * 2 decisions
= 2^3 = 8

but we want to exclude the case where all 3 toppings are not chosen (because we know at least 1 topping IS chosen) so 8-1 = 7

hope that helps
Interesting explanation thanks :)
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