# Combinatorics: Permutation Problems

Consider the word BYTES, how many different 3-permutations can be generated if the first letter must be B.

Solution:

I found this problem to be interesting because it takes a little bit of thinking. You can’t just plug in numbers to any formula to get an answer which makes it the perfect type of quiz or test question.

The first thing to realize is that we are asking for some type of r-permutation, but we have a restriction that the first letter must be ‘B’. Since the first letter must be ‘B’ there is only one choice for that position in our 3-permutation, which leaves two spots lefts. Now we must decide all possible combinations of two letters, but since ‘B’ must be placed at position one, we exclude it from our possible choices. The answer becomes

P(4,2) = $\frac {4!}{(4-2)!} = 12$