SOLUTION: If a man has eight pens. Four are black, three are blue, and one is red. How many different ways can the man choose one pen of each color to put in his pocket and carry around with
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Question 388358: If a man has eight pens. Four are black, three are blue, and one is red. How many different ways can the man choose one pen of each color to put in his pocket and carry around with him?
b)how many ways can the man line up or arrange the eight pens on his desk? Answer by gotpork1(3) (Show Source):
a)How many different ways can the man choose one pen of each color to put in his pocket and carry around with him?
Since it asked for 1 of each color, we must find the possible ways for each seperate color
Black: 4P1
Blue: 3P1
Red: 1P1
Since the events are non mutually exclusive, we need to find the solutions for black AND blue AND red:
=(4P1)*(3P1)*(1P1)
=(4)*(3)*(1)
=12
b)how many ways can the man line up or arrange the eight pens on his desk?
Because the 8 pens are not unique elements, and we are looking for the amount of unique ways to arrange, we have to consider how many of each color there are in our solution.
total ways of arranging : = 8! or 40320
ways of arranging red: 1
ways of arranging blue: 3! or 6
ways of arranging black: 4! or 24
So when we are writing the final equation, we get something like this
(