In a random arrangement of the letters in the word GEOMETRY, what is the probability that the two E’s will be consecutive?
Solution to yesterday’s problem:
To illustrate permutations, we can use blanks like this:
____ ____ ____ ____ ____ ____ ____
Each blank represents a letter in the word ALGEBRA. Since there are 7 letters, there are seven possible letters to put in the first blank (we’ll sort out complications afterward):
__7__ ____ ____ ____ ____ ____ ____
There would be seven possible letters for the second blank, but one of them has already been used in the first blank, so there are six. This pattern follows for the rest of the blanks.
__7__ __6__ __5__ __4__ __3__ __2__ __1__
So to find the number of permutations (arrangements of the letters), we just multiply these numbers together to get 5040.
But wait, there’s more! Scrutinizing the word ALGEBRA, we notice that there are two A’s. This is the only case in this problem where order does not matter. For instance, in the rearrangement LGBERAA, it doesn’t matter if the first A came from the beginning of the original word or not. Fortunately, this is easy to fix—simply divide by 2 (one where the first A came from the beginning, and one where it didn’t). For other problems, you would divide by the factorial of the number of times the letter occurs (6 for 3 A’s, 24 for 4 A’s, etc.)
Dividing by 2, we find that there are 2520 arrangements for the letters in the word ALGEBRA.