Exercise 2 Page 713

We were told that a bag contains one green marble, one red marble, and one blue marble. We were also given the outcomes of randomly drawing three marbles from the bag without replacement.

We found how many combinations and permutations of the three marbles can be drawn from the bag.

	Combinations	Permutations
Formula	_3C_3 = 3!/(3-3)!*3!	_3P_3 = 3!/(3-3)!
Simplify	1	6

We are asked to describe a relation between these two values. We can see that the number of combinations is equal to the number of permutations divided by 3 factorial, which is 6. Now, let's take a look at the formulas for the combinations and permutations of n objects taken r at a time. ccc Combinations && Permutations [0.5em] _nC_r = n!/( n- r)!* r! && _nP_r = n!/( n- r)! We can see that we the only difference between them is that the formula for combinations has an additional multiplication in the denominator. We will rearrange the formula for permutations to determine a relation. Let's do it! We can also write this relation in terms of the permutations. _nP_r = _nC_r* r!