So given that there are 26 alphabets in the English language, how many possible three letter words are there? And we're not going to be Now let's ask some interesting questions. T, U, V, W, X, Y, and Z,' you'll get, you'll get 26, 26 alphabets. So if you go, 'A, B, C,ĭ, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, And in case you don't remember and are in the mood to count, So let's ask ourselves some interesting questions aboutĪlphabets in the English language. So now you will have to erase 5 out of every 6 entries, and the number of entries will become /6 = (10!)/(3!*2!). These will ALL become J-E-N-N-Y-J-I-A-N-G, and likewise every single entry will be repeated 6 times when N1, N2, and N3 are all changed to N, since there are 3! ways to put the three 'different' N's into three spots. When you look through the list now, you'll see that every entry will show up exactly 3! (= 6) times: Now, with your list down to (10!)/2 (= 1814400) entries, imagine you decide the same about the N's: N1 and N2 and N3 all get changed to just N. For example, the pair above would BOTH become: J-E-N1-N2-Y-J-I-A-N3-G, and you would need to go through and erase exactly HALF of the permutations, so you would be at (10!)/2. You would find that in your list, suddenly every entry would be in the list TWICE, once from using J1 then J2, and another from using J2 then J1. Now, imagine that ENTIRE list of 10! (= 3628800) permutations (maybe you paid your little brother to write the entire list :) ), and imagine deciding, 'Oh, actually, J1 and J2 should both just be J'. In that case, like you said, there would be 10! different permutations of the 10 letters, since, for example, J1-E-N1-N2-Y-J2-I-A-N3-G and J2-E-N1-N2-Y-J1-I-A-N3-G would be different (I switched the J1 and J2). That is, your name was spelled J1-E-N1-N2-Y-J2-I-A-N3-G, so that there were 10 'different' letters in your name. Let's say for a moment that the J's and the N's were labelled, so we thought of them as different.

Permutation math