Expert: Josh - 2/16/2010

Question
1. How many ways can 6 different foods be arranged in 6 dishes around a lazy Margaret?
2. How many different 5 card poker hands can be dealt from a standard deck of 52?
3. A University offers 5 science courses, 6 humanity courses & 3 literature courses. In how many ways can a student choose 2 science courses, 3 humanity courses & 1 literature course?

Answer
Angela,

Q1. If I understood correctly, the plates are arranged in a circle. To figure out the answer, let us visualize what happens as we fill each plate with food, one by one.

In a "permutations on a circle" type of problem, it does not matter where we begin. Initially, there is not a single location that we can make a reference to. All six plates appear identical. Therefore, we are not interested in what goes onto the first plate, for the simple reason that something has to be placed there, for us to be able to reference a location. So, the selection on the first plate serves only as a marker.

From this point onward, the placement of all other food items will be relative to the position of this reference plate. e.g., we can identify one plate to its left, two plates to its right etc. Now, the order of selection matters.

Imagine all six plates being labeled "R A B C D E". This forms a ring.

First, we can select 1 from 5 different types of food to fill plate "A".
Next, we can select 1 from 4 remaining types of food to fill plate "B".
Then, we can select 1 from 3 remaining types of food to fill plate "C".
Finally, we can select 1 from 2 remaining types of food to fill plate "D".
Whatever remains is automatically placed on plate "E".

So, the overall task can be done in 5*4*3*2*1=120 ways.

In general, given n distinguishable objects we can arrange them in (n-1)! ways in a circle.
| The factorial notation n! is defined as n*(n-1)*(n-2)*...*2*1.
| For example, 5!=5*4*3*2*1. By definition, 0!=1 ]
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2. This question is of the "choose m from N" type.
Suppose the five cards we end up picking are represented symbolically by "B,C,D,E,F".
- "B" must come from 1 of 52 cards.
- Selecting without replacement, the next card "C" must come from 1 of 51 cards remaining.
- Repeating this argument, the next card "D" must come from 1 of 50 cards remaining.
- The next card "E" must come from 1 of 49 cards remaining.
- Finally, the card "F" must come from 1 of 48 cards remaining.
However, since the order in which the cards are drawn does not matter, these five cards can come about in 5*4*3*2*1=120 ways which are indistinguishable.
|
| Illustration:
| Two of the 120 possible realizations are "B,D,E,C,F" and "C,F,B,D,E".
| Although they are selected in a different order, we are still dealt
| with the same hand.
|
So, we have over-counted by a factor of 120.

The total number of combination is in fact 52*51*50*49*48/(5*4*3*2*1).
This is equivalent to C(N,m)=N!/[m!(N-m)!], where N=52, m=5.

In general, there are C(N,m) combination to the "choose m from N" problem, where order does not matter.
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3. We multiply together the number of ways calculated for each category.
Science course: choose 2 from 5 (order does not matter) => C(5,2)=5*4/(2*1)=10
Humanity course: choose 3 from 6 (again, order not matter) => C(6,3)=6*5*4/(3*2*1)=20
Literature course: choose 1 from 3 => C(3,1)=3.

Altogether, the required pattern of courses can be selected in 10*20*3=600 ways.