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Permutations and Combinations

Module by: Don Johnson

Summary: Discusses the basics of combinations and permutations, and how to calculate the probability of certain events, such as n-bit errors in a codeword.

Permutations and Combinations

The lottery "game" consists of picking kk numbers from a pool of nn. For example, you select 66 numbers out of 6060. To win, the order in which you pick the numbers doesn't matter; you only have to choose the right set of 66 numbers. The chances of winning equal the number of different length-kk sequences that can be chosen. A related, but different, problem is selecting the batting lineup for a baseball team. Now the order matters, and many more choices are possible than when order does not matter.

Answering such questions occurs in many applications beyond games. In digital communications, for example, you might ask how many possible double-bit errors can occur in a codeword. Numbering the bit positions from 11 to NN, the answer is the same as the lottery problem with k=6k6. Solving these kind of problems amounts to understanding permutations - the number of ways of choosing things when order matters as in baseball lineups - and combinations - the number of ways of choosing things when order does not matter as in lotteries and bit errors.

Calculating permutations is the easiest. If we are to pick kk numbers from a pool of nn, we have nn choices for the first one. For the second choice, we have n-1 n 1 . The number of length-two ordered sequences is therefore be nn-1 n n 1 . Continuing to choose until we make kk choices means the number of permutations is nn-1n-2…n-k+1 n n 1 n 2 … n k 1 . This result can be written in terms of factorials as n!n-k! n n k , with n!=nn-1n-2… 1 n n n 1 n 2 … 1 . For mathematical convenience, we define 0!=1 0 1 .

When order does not matter, the number of combinations equals the number of permutations divided by the number of orderings. The number of ways a pool of kk things can be ordered equals k! k. Thus, once we choose the nine starters for our baseball game, we have 9!=362,880 9 362,880 different lineups! The symbol for the combination of kk things drawn from a pool of nn is nk n k and equals n!n-k!k! n n k k .

Exercise 1

What are the chances of winning the lottery? Assume you pick 66 numbers from the numbers 11-6060.

Solution 1

606=60!54!6!=50,063,860 60 6 60 54 6 50,063,860 .

Combinatorials occur in interesting places. For example, Newton derived that the nn-th power of a sum obeyed the formula x+yn=n0xn+n1xn-1y+n2xn-2y2+…+nnyn x y n n 0 x n n 1 x n 1 y n 2 x n 2 y 2 … n n y n .

Exercise 2

What does the sum of binomial coefficients equal? In other words, what is ∑k=0nnk k 0 n n k

Solution 2

Because of Newton's binomial theorem, the sum equals 1+1n=2n 1 1 n 2 n .

A related problem is calculating the probability that any two bits are in error in a length-nn codeword when pp is the probability of any bit being in error. The probability of any particular two-bit error sequence is p21-pn-2 p 2 1 p n 2 . The probability of a two-bit error occurring anywhere equals this probability times the number of combinations: n2p21-pn-2 n 2 p 2 1 p n 2 . Note that the probability that zero or one or two, etc. errors occurring must be one; in other words, something must happen to the codeword! That means that we must have n01-pn+n1p1-pn-1+n2p21-pn-2+…+nnpn=1 n 0 1 p n n 1 p 1 p n 1 n 2 p2 1 p n 2 … n n p n 1 . Can you prove this?

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