Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

Letters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

Can anyone please explain why we can't count the probability like this?:

5/9 x 4/8 x 4/7 x 3/6 (VxVxCxC) = 5/63

what is the meaning of ^ sign here?

I'm not sure if you are talking about the AUSTRALIA problem. In that problem we need a word that have 4 letters, and for each letter there are 9 possible choices, so the total outcome is 9*9*9*9=9^4.


This is a very good question. In this question each secretary can have 0 to 4 reports to type, but each report must be and can only be typed by one person. Here you need to look which one is replaced back when you repeat the test. If a report is typed by one person, then it will not be typed by another person. In other words, the reports can not be replaced. On the other hand, if a secretary has typed one report, she'll still be placed in the pool for the second report. In other words, the secretary is the one that gets replaced when we repeat the test, and each report faces the same pool of three secretaries. Therefore, the outcome is 3*3*3*3 for the four reports.

Let me give you another example, see if it will help.

Example:
1. Five rooms are to be allocated to four people. How many ways are there if each room must be assigned to one person and one person only?
Here each people can have more than one room, or have no room at all. In other words the people will be replaced back into the pool in each repeated test and each room will face the same pool of people. So five room each face a pool of four people. 4*4*4*4*4=4^5

2. Five rooms are to be allocated to four people. How many ways are there if each person must be assigned to one room and one room only?
Here each room can be empty, or can have multiple people assigned to it. In other words the rooms will be placed back into the pool for each repeated test and each person will face same pool of five rooms. Therefore total outcome is 5*5*5*5=5^4.

Guys, please help me with the following problems:

1. A committee of 3 people is to be chosen from four 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A.28 B.32 C.34 D.41 E.56

The way you are doing it, you've given it a specific order.

For example, if there are 2 red balls and 2 white balls, what is the probability of getting 1 white ball and 1 red ball out in the first two draws? It's C(2,1)*C(2,1)/C(4,2) = 2/3.
You could write it out to verify:
R1R2
R1W1
R1W2
R2R1
R2W1
R2W2
W1R1
W1R2
W1W2
W2R1
W2R2
W2W1

But using your method you will have 2/4*2/3=1/3. Your problem is you have specified that you will draw a white ball first, and then red ball second. However there is also the possibility of drawing a red ball first, and then a white ball second. So you need to add another 1/3 to your result to get 2/3.