Combinatorial functions

If n is a positive integer then n factorial, written n! is defined as

We also define 0! = 1.

If m and n are non-negative integers with mn, the combination function C(n, m) is the number of ways m different elements can be selected from a set consisting of n different elements. This function can be expressed as

If m and n are non-negative integers with mn, the permutation function P(n, m) is the number of ways of ordering m different elements selected from a set consisting of n different elements. This function can be expressed as

Note that the following is true:

Example 1:

1. How many different ways are there to order the letters in the word GRAPH?
2. How many three letter “words” can be formed from the letters in GRAPH without repeating any letter?
3. How many ways are there to draw three letters from the letters in GRAPH (without worrying about their order)?

The problem is equivalent to finding the values of (a) 5!, (b) P(5,3), and (c) C(5,3).

4 Responses to Combinatorial functions

1. jody says:

Do you know of a way in MS-Excel to compute 5!

Thanks

• Charles says:

=FACT(5)

2. Peter Chase says:

I have a knowable stats question, but one that I can’t do by brute force. Thanks for your help!

Is there an excel plug in / online formula for: probability of X unrelated outcomes given N events, each with set probabilities?

As in:
P(event1)=70%
P(event2)=50%
P(event3)=60%
P(event4)=80%
P(event5)=40%
P(event6)=90%

so…what’s P (one event occurs; don’t care if it’s 1, 2,3,4,5, or 6)
P (2 events occur; don’t care which ones)
P (3 event occur; don’t care which ones)
P (4 events occur;)
etc

• Charles says:

Peter,
Actually you haven’t provided enough information to determine the answer. You need to know things like P(E1 and E2), etc. E.g., suppose you put 10 balls numbered 1 to 10 in a bag and pick one of them at random. In Scenario 1, E3 is ball 1, 2, 3, 4, 5 or 6 is picked and E5 is that ball 7, 8, 9 or 10 is picked. Then P(one event occurs) = 100% (irrespective of the definitions of the other events). In Scenario 2; E1 is ball 1, 2, 3, 4, 5, 6 or 7 is picked; E2 is ball 1, 2, 3, 4 or 5 is picked; E3 is ball 1, 2, 3, 4, 5 or 6 is picked; E4 is ball 1, 2, 3, 4, 5, 6, 7 or 8 is picked; E5 is ball 1, 2, 3 or 4 is picked; E6 is ball 1, 2, 3, 4, 5, 6, 7, 8 or 9 is picked. Then P(one event occurs) = 90%.
Charles