Page 11 - Intro to Tensor Calculus

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EXAMPLE 1.1-4. The total number of possible ways of arranging the digits 1 2 3 is six. We have
three choices for the first digit. Having chosen the first digit, there are only two choices left for the second
digit. Hence the remaining number is for the last digit. The product (3)(2)(1) = 3! = 6 is the number of
permutations of the digits 1, 2 and 3. These six permutations are

1 2 3 even permutation
1 3 2 odd permutation

3 1 2 even permutation
3 2 1 odd permutation

2 3 1 even permutation
2 1 3 odd permutation.

Here a permutation of 1 2 3 is called even or odd depending upon whether there is an even or odd number
of transpositions of the digits. A mnemonic device to remember the even and odd permutations of 123
is illustrated in the figure 1.1-1. Note that even permutations of 123 are obtained by selecting any three
consecutive numbers from the sequence 123123 and the odd permutations result by selecting any three
consecutive numbers from the sequence 321321.

Figure 1.1-1. Permutations of 123.

In general, the number of permutations of n things taken m at a time is given by the relation

P(n, m)= n(n − 1)(n − 2) ··· (n − m +1).

By selecting a subset of m objects from a collection of n objects, m ≤ n, without regard to the ordering is
called a combination of n objects taken m at a time. For example, combinations of 3 numbers taken from
the set {1, 2, 3, 4} are (123), (124), (134), (234). Note that ordering of a combination is not considered. That
is, the permutations (123), (132), (231), (213), (312), (321) are considered equal. In general, the number of
n n!
combinations of n objects taken m at a time is given by C(n, m)= = where n are the
m m!(n − m)! m
binomial coefficients which occur in the expansion

n
X n
n
(a + b) = a n−m m
b .
m
m=0