I'm going to try to collect some P and C type questions and approaches in this thread for easy reference for everybody. Please feel free to discuss and add to the thread.

Repeated Outcomes

If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

Explanation: For each object, there are k outcomes. So the total number of outcomes would be k*k*k ...*k, for n times.

Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total number of fruits = 9
For each fruit there are 2 possible outcomes: included, not included
Total outcome = 2^9-1
(The minus one is to take out the one possible outcome where nothing is in the basket.)

Example:

Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total outcome = (2^3-1)*(2^4-1)*(2^2-1)

Example:

There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^3=27
Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6
Probability = 6/27=2/9

Example:

There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?

For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C
Total outcome: 3^4=81
Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36
Probability = 36/81=4/9

Question: in the above example, why isn't total outcome = 4^3?

The way I approach i took is.. whenever I see atleast i go ahead using 1 minus..

Not necessarily. It depends on how convenient either approach is. For example, if asked the probability that among ten babies at least two are boys, I'd use 1 minus the probability of zero boy and one boy, but if asked the probability that among ten babies at least nine are boys, I'd simply add up the probabilities of nine boys and ten boys.

makes sense..thanks guys.

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

Example:
From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?

This is a case of "with replacement":
Total outcomes: 9^4
Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2
You can calculate the probability from here.

Note that there is a special thing with letter problems, ie. the repeating letters. In the above examples the repeating letters don't matter. However they would matter if the question is asked differently.

Example:
There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?

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

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)

After you've got the four letters you need to order them to get different outcomes. So the first case (with different vowels) would be C(3,2)*C(4,2)*P(4,4)=432, and the second case (with same vowels) would be C(1,1)*C(4,2)*P(4,4)/2=72. The final outcome would be 504.

For Outcomes with two vowels and two consonants, why isn't it: "4!" instead of P(4,2)?




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From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?

This is a case of "with replacement":
Total outcomes: 9^4
Outcomes with two vowels and two consonants: P(4,2)*5^2*4^2
You can calculate the probability from here.

Question: in the above example, why isn't total outcome = 4^3?

Hong in the 3 secretaries and 4 departments problem why di we get the solution as 3^4 and not 4^3? Please explain the concept

Thanks

How can I count C(n,k)?
Thank you

( Remember this question )

Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible?
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Its not the fact its your attitude towards the fact