Why Do We Multiply Combinations in Probability Questions?

[Peter G.]

Hi,

I have a terrible time understanding one of the concepts in the following question:

How many 7 persons committees consisting of 4 females and 3 males may be assembled from a pool of 17 females and 12 males.

Ok, so I get this is a combination question.

Splitting this between male and females:

Male: (12 x 11 x 10)/(3!)

Female: (17 x 16 x 15 x 14)/(4!)

Why, however, do we have to multiply the number of combinations by each other to get the final result? Why not add?

Thanks in advance,

Peter

 

[Ray Vickson]

Hi,

I have a terrible time understanding one of the concepts in the following question:

How many 7 persons committees consisting of 4 females and 3 males may be assembled from a pool of 17 females and 12 males.

Ok, so I get this is a combination question.

Splitting this between male and females:

Male: (12 x 11 x 10)/(3!)

Female: (17 x 16 x 15 x 14)/(4!)

Why, however, do we have to multiply the number of combinations by each other to get the final result? Why not add?

Thanks in advance,

Peter

You can answer this question for yourself (and doing so is the BEST way to learn); just make the example smaller: choose a committee of 1 male and 1 female from a group of 2 males and 3 females. That example is small enough that it allows you to list all the possibilities and then count them. You will see right away whether you should add or should multiply.

RGV

 

[HallsofIvy]

The "fundamental counting principle": If one thing can happen in n ways and another thing can happen, independently of the first, in m ways, then both can happen in n times m ways.

There are 12 males so there are 12 ways to choose the first male. There are then 11 males left so there are 11 ways to choose the second. There are then 10 males left so there are 10 ways to choose the third. By the "fundamental counting principle", there are 12(11)(10) ways to choose 3 males form 12 in that order. Since order is not important and there are 3! different orderings, there are 12(11)(10)/3! ways to choose the three males.

There are 17 females so there are 17(16)(15)(14)/4! ways to choose four of them.

Then, by the "fundamental counting principle" again, there are [12(11)(10)][17(16)(15)(14)]/(3!4!) ways to choose 3 males and 4 females.

 

[Villyer]

A logical shortcut with probability is wording; if the question mentions 'and' then it implies multiplication, if it says 'or' it implies addition. This isn't always true, but it can still help.