Video Transcript
If you randomly select a permutation of the letters shown, what is the probability that they spell “kinetics”?
A quick glance at the letters shows that we do have all the right letters for spelling the word kinetics. The question is if these letters are randomly permuted, that is arranged, what is the probability that they’re in the right order to spell the word kinetics. We need to think about how you calculate the number of different permutations of a group of objects. There is an extra complication here in that the letters are not all distinct; there’re two eyes and we must take this into account. Here’s the key rule that we need: the number of distinct permutations of 𝑛 objects in which one object is repeated 𝑟 times is 𝑛 factorial over 𝑟 factorial. Remember, a factorial is found by multiplying together all of the integers from one to that number.
So let’s determine the values of 𝑛 and 𝑟 for this question. 𝑛 is the number of objects, so that’s the number of letters which in this case is eight. 𝑟 is the number of times that a particular letter is repeated. So here we have the letter 𝐼 and it appears twice. So 𝐼 is equal to two. If in a different problem we had for example three Ks, we would also divide by three factorial to account for this. But as I is the only repeated letter here, we just divide by two factorial. So the total number of permutations of the given letters is eight factorial divided by two factorial. Eight factorial is 40320 and two factorial is just two. So we have the fraction 40320 over two. This is equal to 20160.
So we know the total number of permutations of the given letters. However, the question asked us for the probability that the given permutation forms the correct spelling of the word kinetics. So we need to convert this answer into a probability. Well, only one of these 20160 permutations forms the correct spelling of the word kinetics.
Therefore, the probability that we randomly select the correct permutation out of the 20160 possibilities is just one out of 20160.
