Big Ideas Math Geometry, 2014
BI
Big Ideas Math Geometry, 2014 View details
5. Permutations and Combinations
Continue to next subchapter

Exercise 52 Page 706

Practice makes perfect
a A bag contains a green marble, a blue marble, and a red marble. We want to know how many combinations of 3 marbles are there. Let's write the formula to find the combinations of n objects taken r at a time.
_nC_r = n!/( n- r)!* r! We will find the combinations of 3 objects taken 3 at a time.
_nC_r = n!/(n-r)!* r!
_3C_3 = 3!/( 3- 3)!* 3!
_3C_3 = 1/(3-3)!
_3C_3 = 1/0!

0!=1

_3C_3 = 1
There is only one combination because the order in which we draw the three marbles does not matter. In the end, we will always have one green marble, one red marble, and one blue marble.
b A bag contains a green marble, a blue marble, and a red marble. We want to know how many permutations of 3 marbles are there. Let's write the formula to find the permutations of n objects taken r at a time.
_nP_r = n!/( n- r)!We will find the permutations of 3 objects taken 3 at a time.
_nP_r = n!/(n-r)!
_3P_3 = 3!/( 3- 3)!
_3P_3 = 3!/0!

0!=1

_3P_3 = 3!

Write as a product

_3P_3 = 3*2*1
_3P_3 = 6
Let's look at every possible outcomes of drawing three marbles from the bag.

Since the order matters in a permutation, each outcome is associated with a different permutation. We can see that there are 6 different permutations and, therefore, there are 6 different outcomes.