Permutations and factorials

Definition. We call permutations all possible arrangements of ndifferent objects on a row.

Definition. If n is a natural number, then n! (pronounced ``n factorial'') is the product of $1\cdot2\cdot3\cdot...\cdot n$. Therefore 1!=1, 2!=2, 3!=6, etc. For convenience and for consistency 0! is defined to be equal to 1.

Methodological Remark. Before working with permutations one must know the definition of factorial and learn how to deal with this function. The following exercises might be useful.

1.

Simplify the expressions: a) 10!.11; b) n!.(n+1).

2.

a) Calculate 100!/98!; b) Simplify n!/(n-1)!.

3.

Prove that if p is a prime number, then (p-1) is not divisible by p.

Problem7. How many ways are there to lay four balls colored red, black, blue and green in a row.
Solution. The first place in the row can be occupied by any of the given balls. The second can be occupied by any other remaining three balls at cetera. Finally, we have the answer: $4\cdot3\cdot2\cdot1=4!$

Similarly, we can prove that n different objects can be laid out in a row in exactly $n\cdot(n-1)\cdot(n-2)\cdot...\cdot2\cdot1$ ways, i.e,

Theorem. The number of the permutations of n objects is n!.

In the next 2 problems we will call ``word'' any combination of letters.

Problem8. How many ``words'' can be created by permuting the letters of the word ``VECTOR''
Solution. Since this word consists of 6 different letters, the answer is 6!

Problem9. How many ``words'' can be created by permuting the letters of the word ``CARAVAN''
Solution. Thinking of the letters A in this word as A₁, A₂, AND A₃, we get 8! different words. However, any words which can de obtained from each other by transposing the letters A_i are identical. Since the letters A_ican be rearranged in 3! ways, all 8! words split into groups of 3! identical words. Therefore the answer is 8!/3!.

Problem10. How many diagonal has got a convex n-gon?
Answer. n(n-3)/2