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July 20th, 2013, 12:48 PM   #1
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PERMUTATION AND COMBINATIONS

Hey guys, need help on solving question (b) so I know how to approach a similar question in the exam.

(a) If the options are independent, how many different combinations of options and blends exist:

- Regular / Large
- Black / White / Cream
- Strong / Medium / Weak
- Coffee brand: 1 / 2 / 3 / 4 / 5 / 6 (6 different coffee brands to 1 choose from)

I worked this out by using the formula (N! / (N-K)!K!) for each option and multiplied the outcome together: 2x3x3x6 = 108 different combinations .

(b) If a customer buys three coffees, all different, how many possible orders are there? Explain the reasoning behind your answer as well as giving a method and a numerical result.

How would you guys go about this?
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July 20th, 2013, 01:11 PM   #2
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Re: PERMUTATION AND COMBINATIONS

108x107x106/1x2x3 I'm sure you can do the arithmetic.
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July 20th, 2013, 03:11 PM   #3
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Re: PERMUTATION AND COMBINATIONS

Quote:
Originally Posted by mathman
108x107x106/1x2x3 I'm sure you can do the arithmetic.
204156?
I didn't think the combination would be more than how many options were available though.
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July 21st, 2013, 12:16 PM   #4
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Re: PERMUTATION AND COMBINATIONS

For the first coffee there are 108 choices, for the second (different) there are 107, and for the third 106. Since the choice order doesn't matter divide by 3!.
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July 21st, 2013, 02:06 PM   #5
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Re: PERMUTATION AND COMBINATIONS

Hello, maths4computing!

Quote:
204,156?
I didn't think the combination would be more than how many options were available though.

You're kidding, right?


A menu offers 10 items:
[color=beige]. . [/color]2 appetizers, 5 entrees, 3 desserts.

Choose one from each category,
[color=beige]. . [/color]there are: 2 x 5 x 3 = 30 possible dinners.


There are only 52 cards in a standard deck of cards.
But there are 2,598,960 possible 5-card poker hands.

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