Question

PLEASE HELP!!!!

Logan is curious as to how many ways he can rearrange the letters in his name.
Assuming that n must be the last letter, how many ways can Logan arrange the the letters in his name?

Please also explain how you'd get the answer to this problem. Thanks! :)

Answer 1

Well, if we were to arrange the letters in such a way that we may have any letter anywhere, we have 5 letters that can go into the first spot, 4 letters that can go into the second spot (since one letter goes into first spot already), 3 letters that can go into the third spot, 2 letters that can go into fourth spot, and then only 1 in the last spot. This gives 5*4*3*2*1, or 5! ways to arrange the letters.

However, since we made "n" the last letter, we can do the same thing only with the first four letters (since the fifth is just not going to change anyways)

Answer 2

Perfect! Thanks so much, AccessDenied! :)

Answer 3

You're welcome! You get 4!, correct? :)

Answer 4

Wait - I'm not sure what you mean. Are you saying that, because we have only 1 choice for the last spot, that would give us this: 4, 3, 2, 1, 1 ?

Answer 5

Since the "n" is already a given (by the problem), it's not going to move around anyways. We can only consider what the first four letters will do.
Like this: _ _ _ _ N: L, O, G, and A can move around in those four spaces.
4 for the first spot, 3 for the second, 2 for the third, and just one for the last spot.

Answer 6

Ah - okay! I see, now! Yes, I do see how you get 4! Thanks so much for the thorough explanation!

Answer 7

You're welcome! Yeah, in general, when we're arranging letters in a word or any string of letters really, we can find the number of possible arrangements.
If we have \(n\) letters (no repeating letters), then there is \(n!\) ways to arrange them.

Just a note, though, that if we have repeating letters, we have to divide by the number of arrangements that each letter can make alone to remove the duplicates. Like, if the name were like "Anna," we would have \(4! / (2! \times 2!)\) dividing off the duplicate arrangements made by the two "A"s and "N"s.

Answer 8

Perfect! That helps a lot! Probability is always tough. I'll remember this, the next time I have a similar type of problem. Thanks for explaining this so well! It is very easy to understand. :)

Answer 9

Definitely is difficult (I think). I honestly just got the intuition today on why you divided by the number of arrangements for repeats. Since there are some number of arrangements that are all duplicates, we divide it off from the total because it essentially makes it "1,"

Answer 10

Ah . . . well good for you for figuring that out! I love those moments where it just comes to you! I think it makes sense, too.