# Probabilities When we use the Combinations and when not? nCk vs nPk

Question 1:

I am working on probabilities and on some exercises the solutions either use the nCk or just n. I want to find a method to understand when I have to use the nCk and when not to.

Example 1 (nCk is used):

What is the probability that a five-card poker hand contains the two of diamonds, the three of spades, the six of hearts, then ten of clubs, and the king of hearts?

Solution: 1/ 52 C 5


Example 2 (nCk is not used):

What is the probability that a five-card poker hand does not contain the queen of hearts?

Solution: 47/52


Two similar questions with different Sample Space! Is any tip or standard rule that can be used to determine whether a combination or Permutation must be used, and when just the number of all possibles events (e.g S=52) ?

Question 2:

Some exercises do not determine which method must be used (Combinations or Permutations). As I know, Combinations is used when order doesn't matters and repetition is not allowed, on the other hand Permutations is used when order matters and repetition is not allowed. Both have another kind of method where the repetition is allowed (n+k,C,k and n^k respectively). Any tip or trick when should I use the Combinations or Permutations? and when to use the unlimited repetition for both? I knew that exercise must make it clear, but I find it a bit hard to deal with this choice of the method.

You can use even permutations in you first question! Why not? It will be just multiplying $5!$ in both numerator and denominator which just get cancelled out!

Also in the second case : $\dfrac{^{51}C_5}{^{52}C_5}=\frac{47}{52}$

First represents choosing $5$ cards from pack without the heart queen and denominator from all cards.

Both give same result.

And generally the question states what to do. If it is unclear, just do what you feel easier. Because people forget to explain elementary cases. But, still, the question should make it clear.

• Thank you! In general, when I should use the permutations the exercise will define that order matters? And it is more frequently that when the exercise do not state that order matters, is like the order doesn't matters? Can I use this logic to understand the problem? Commented May 18, 2014 at 17:38
• @Dr.Mathematician In most cases it is governed by common sense. Like your cards are distinct. Taking them to be distinct prevails in most cases. Commented May 18, 2014 at 17:42
• Because English is my second language, it is possible to list this words here and what is meant by that word (what should i use) as I knew (not sure) distinct means with out repetition? If it is not a pain for you, I will be very pleasure if you list them on your original post, which will help me for my exams but also for other users. Cheers again! Commented May 18, 2014 at 17:54
• @Dr.Mathematician English is also my second language! Cheers! It's just practice after some like 30 different questions you will be well versed with the terminology. Commented May 18, 2014 at 17:58

Combinations are just another name for "set". Order of the elements is irrelevant.

Permutations are arrangements where order is important.

This should cover most cases. Two general rules (from which the above can be derived) is that if you have a sequence of decisions (like selecting each element in a cartesian product of sets) you multiply. If you have a union of disjoint sets, add sizes of the sets to get the total size. Sometimes you are including elements twice, subtract them. It is sometimes easier to divide into a collection of disjoint sets that are easy to handle.

Look if it is easier to use the sum rule backwards: Compute the total, subtract those that don't satisfy the restrictions.

Dr Mathematician$$P(x)=nCr\cdot (p^r)\cdot \left (q^{n-r}\right)$$ is normally used when you have a case of "Success" and "Failure". The formula is $$P(x)=nCr\cdot (p^r)\cdot \left (q^{n-r}\right)$$ In Q1 they really specify 2 of Diamon, 3 of Spades, 6 of Heart, 10 of club and king of heart. So you will be calculating probability of each card, then you will have to sum it up in the end. In this case each card will be of 20%, therefor change to 0.20 and it will be equal to p and q automatically will be equal to 0.80 Q1 they give you five card which is equal to 100%, then they give you Question 1 is a type of Binomial Distribution. This should work. ;-)

Notice nPr is a Permutation Probability Concepts.