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by smallsorrow » Mon Oct 27, 2008 4:11 am
The lunch menu contains 4 different entrees and 5 side dishes. A meal consists of 1 entree and 2 side dishes. What is the number of combo?

IMO 4x5x5

But that is wrong can anyone explain why?

Cheers, Lisa
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by stop@800 » Mon Oct 27, 2008 4:40 am
Number of combos will be
4C1 * 5C2
= 20

In your case
5*5 is not correct for two reasons

1: If dish1 has been choosen than for second dish it has to be 4 options
so it will be 5*4

2: "Dish1 in first selection and Dish2 in second selection" and
"Dish2 in first selection and Dish1 in second selection" are same.

In other words, we need to use combination because order does not matter.

so we need to divide the whole by 2
ie 5*4 / 2
=10

so ans will be
4*10
and not
4*5*5

Hope this helps!!
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by KeyserSoze525 » Mon Oct 27, 2008 4:42 am
The question says there are 4 Entrees and 5 Sides. A meal consists of 1 Entree and 2 Sides.

So the setup would be:

_ _ _
E S S

For the Entrees, you would calculate 4 choose 1 or 4C1 = 4!/1!*(4-1)! = 4

For the Sides, you would calculate 5 choose 2 or 5C2 = 5!/2!*(5-2)! = 10

So the answer is 4*10 = 40
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by stop@800 » Mon Oct 27, 2008 4:54 am
stop@800 wrote:Number of combos will be
4C1 * 5C2
= 20
typo
4C1 * 5C2 = 40
and not 20 :)
In your case
5*5 is not correct for two reasons

1: If dish1 has been choosen than for second dish it has to be 4 options
so it will be 5*4

2: "Dish1 in first selection and Dish2 in second selection" and
"Dish2 in first selection and Dish1 in second selection" are same.

In other words, we need to use combination because order does not matter.

so we need to divide the whole by 2
ie 5*4 / 2
=10

so ans will be
4*10
and not
4*5*5

Hope this helps!!
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