1

The total different ways that can the letters in the word

"ARRANGEMENT"

to be arranged in 11 letters is:

$$\frac{11!}{2!\times2!\times2!\times2!}=2494800\space ways $$

But how many different ways can the letters be arranged in only 5 letters?

Ho Chung Yin
  • 31
  • 1
    Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Nov 09 '16 at 11:44
  • 2
    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Nov 09 '16 at 11:44
  • I would also strongly recommend using the search feature. You can often find a similar question like this one which might help you to answer your own question. – Ian Miller Nov 09 '16 at 11:47
  • More than 3 :) , less than there are atoms in the universe – Pieter21 Nov 09 '16 at 11:50
  • @HoChungYin Great first edit. Now what have you tried? And where did you get stuck? – Ian Miller Nov 09 '16 at 12:30
  • @IanMiller I do not know how to calculate the total way of arrangement of 5 letters from the letter in the word "ARRANGEMENT". – Ho Chung Yin Nov 09 '16 at 12:36
  • If it was 11 letters could you do it? If it was 1 letter could you do it? If it was two letters could you do it? Hopefully these questions will let you make a start. – Ian Miller Nov 09 '16 at 12:38
  • @IanMiller I know how to do if it was 11 letters, which is no restriction. But when it come to a restriction, I really don't have any idea to solve it, as it contain too many repeated letters, such as "A","R","E","N". – Ho Chung Yin Nov 09 '16 at 12:46
  • Good. Edit your question and add some of those comments as your thoughts on the question. Then your question can be reopened. I've given it one reopen vote and removed my downvote to get it started. Improve the question and more will follow. – Ian Miller Nov 09 '16 at 12:56

1 Answers1

-2

The $$\frac{11!}{2!^4}$$ part is already right.

To divide it in 5 words, you have to multiply by:

$${{6 + 4}\choose{4}}$$

Given that words have at least 1 letter, you have $11 - 5 = 6$ letters left to divide over 5 buckets for 0 or more letters.

This is typically done with an eggs in baskets calculation with bars: you have to place 4 bars between the eggs (letters in this case). Out of 10 positions, you have to select 4 bars.

Pieter21
  • 3,515