Miller & Freund\'s Probability and statistics, 8th edition, CHAPTER3 PROBLEM 19
ID: 3050313 • Letter: M
Question
Miller & Freund's Probability and statistics, 8th edition, CHAPTER3 PROBLEM 19 One Engineering group consists of 6 men and 4 women. (a) How many different project teams can be formed consisting of of 2 men and 2 women (b). If 2 women have the same boy friend and refuse to be on the same team together, how many different project teams can be formed consisting of 2 men and 2 women?
For this solution is (a) 6C2 . 4C2 =90. It is ok.
But I need a clarification on part (b) solution on Chegg, which is given as "the number of ways 2 women and their boyfriend work in the same team is 15, therefore the no. of different project teams can be formed is 90-15=75". I would like to understand how 15 is arrived at in the solution, Could you please elaborate?
Explanation / Answer
Solution of the part (b):
As per the problem states, among the 4 women 2 have same boy friend and they wish not to work together.
Hence, the only possible way to keep 2 women having same boyfriend in one team is 1, as it will make the other 2 women form a definite another team and there will be no more choices.
But, the number ways to choose 2 men out of the 6 is : 6C2 = 15.
Hence the total number ways 2 women and their boyfriend work in the same team is : 6C2 . 1 = 15
Now as the 2 women don't want to work together, the number of ways different project groups can be formed is nothing but the number of ways complementary event can be formed is:
= total no. of ways - no. of ways 2 women can be togethe
= 90 - 15
= 75
I hope this explaination will be sufficient to help you.
All the best.