SOLUTION: A group of four men and five women. Three people are selected to attend a conference. a. In how many ways can three people be selected from this group of nine? b. In how m

Question 161107This question is from textbook Thinking Mathematically
: A group of four men and five women. Three people are selected to attend a conference.

a. In how many ways can three people be selected from this group of nine?

b. In how many ways can three women be selected from the five women?

c. Find the probability that the selected group will consist of all women.

This is how I answered:

(a) 9!/(9-3)! 3! = 9*8*7*6*5*4*3*2*1/6*3*2*1 cancel out the 6 on the numerator and denominator = 504/6 = 84 This is the answere the book gave as well.
(b) 5!/(5-3)! 3! = 5*4*3*2*1/2*3*2*1 cancel out the 2 in the numeratoer and denominator = 60/6 = 10 agan, this is also the answer the book gives.
(c) this is where I am stuck I tried this 5!/(9-5)! 3! = 5*4*3*2*1/4*3*2*1 cancelling out the 4 in the numerator and the denominator which will leave you
with 5 as a numerator, but if we cancel out the 4 on the denominator that will leave us with 6 which is the wrong answer. The books states that the answer
should be 5/42.

If you can help me to understand how they come up with 5/42 for I would appreciate it. Thanks a million!
KiKiThis question is from textbook Thinking Mathematically

Answer by vleith(2517) (Show Source):
You can put this solution on YOUR website!
You have a 5 in 9 chance of the first person being a woman.
You then have a 4 in 8 chance of the second one being a woman.
Finally, you have a 3 in 7 chance of the third person being a woman.
So the chances are