Associated Topics || Dr. Math Home || Search Dr. Math

### Possible Combinations from Cards Drawn at the Same Time?

```
Date: 04/02/98 at 22:40:48
Subject: Probability

Four cards were placed in a bag, each having a different number
written on it. All of the cards were the same size, shape and color.
If two of the four cards were drawn AT THE SAME TIME out of the bag,
how many possible combinations could be made?

a. 4
b. 6
c. 12
d. 16

Our class thinks the answer is b (6), but our teacher's book says the
answer is c (12). We think the teacher's book is wrong because they
"forgot" the words "at the same time." Can you help us? Also, is there
a formula for solving probability problems?

Thank you,
```

```
Date: 04/03/98 at 07:30:52
From: Doctor Anthony
Subject: Re: Probability

Two objects can be chosen from 4 objects in 6 ways, but the question
then asks how many combinations of numbers could be made. This is a
poorly worded question, because you could interpret the last part in
two ways: namely, the number 5,3 (for example) might be thought
different from 3,5, or you might think these are the same combination.
My preference is the answer you gave, 6.

There are general formulae for these types of problem. You will find
them in textbooks under the headings "Permutations and Combinations."

If you have, say, 10 objects, then the number of arrangements
(permutations) that could be made from 4 of the objects is given by:

P(10,4) = 10*9*8*7  =  5040

This is because you could choose the first object in 10 ways, the
second object in 9 ways, the third in 8 ways and the fourth in 7 ways.
In this question, the ORDER of the objects matters, so that, for
example, swapping the first and second counts as a different
permutation.

Using the factorial notation where say 5! = 5*4*3*2*1 = 120, we could
write:

10!
P(10,4) = -----  = 5040
6!

With COMBINATIONS, we are not concerned with the order of the objects.
The number of combinations of 4 things that could be chosen from 10
things is given by:

10*9*8*7      P(10,4)       10!
C(10,4) = ----------  = --------  =  ------
4*3*2*1          4!        4! 6!

This is because each of the PERMUTATIONS of 4 things could be
rearranged within itself in 4! ways without counting as a different
COMBINATION. So the expression for the number of permutations is too
large by a factor of 4! when we are trying to find the number of
combinations.

This means we have the formula for C(10,4) as:

10!
C(10,4) = -------
4! 6!

The question you started with was how many ways can we can choose 2
objects from 4.

4!
C(4,2) = ------   =  6
2! 2!

-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations
Middle School Word Problems

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search