How many words with or without meaning can be made from the letters of the word CONCENTRATIONS by taking $4$ letters at a time?

There are $14$ letters total present in the word. $4$ letters can be picked in $14\choose 4$ ways. $4$ letters can be arranged in $4!$ ways. So ideally answer has to be ${14\choose 4} * 4!$. But I know this answer is wrong because some letters are repeating. How do I solve this then? I appreciate any help.

  • 2
    $\begingroup$ Just a note, ${14 \choose 4}$ is the same thing as $14!/4!(14-4)!$, So when you multiply by 4! you are really calculating $14! /10!$. $\endgroup$
    – h94
    Dec 20, 2016 at 18:49
  • $\begingroup$ Answer is $4436$ $\endgroup$
    – Kiran
    Dec 20, 2016 at 18:50

2 Answers 2


See a similar question here. You can search for "number of arrangements using letters of word" in this website and find many similar problems here.

You can also use this tool where you can enter any word and it generates such questions and solutions. For the word CONCENTRATIONS, it gave answer as 4436 which is given below

enter image description here

  • $\begingroup$ sorry, i thought it will be useful for him. can delete, if it is a violation. pl let me know. i find this tool useful while practicing similar questions and hence posted. $\endgroup$
    – Kiran
    Dec 20, 2016 at 19:36
  • $\begingroup$ thanks. is it possible for me to now move my answer to "community wiki" ? any moderators, please help $\endgroup$
    – Kiran
    Dec 20, 2016 at 19:44
  • $\begingroup$ Yes you can: click on edit (to the lower left of your post; that allows you to edit the answer field, look for a very small box that you can click, and to the right of the little box, you,ll see "make this community wiki" (at the very least, you'll see "community wik" to the right of the box, and clicking the box makes the post a "community wiki" post. $\endgroup$
    – amWhy
    Dec 20, 2016 at 19:47
  • $\begingroup$ i did so, thanks $\endgroup$
    – Kiran
    Dec 20, 2016 at 19:49
  • 1
    $\begingroup$ Thanks for you contribution, @Kiran ! :-) $\endgroup$
    – amWhy
    Dec 20, 2016 at 19:52

There are $9$ unique letters: C, O, N, E, T, R, A, I, S. Of those letters, N has multiplicity $3$ (meaning it appears three times), C, T, and O have multiplicity $2$, and all other letters appear only once.

Here's a lengthy, but elementary approach. Any $4$ letter word you make here has one of the following forms:

1) $4$ unique letters, e.g. CONE

2) A pair of same letters, and $2$ distinct letters (distinct from each other and the pair), e.g. TOTE

3) Two pairs of distinct letters e.g. COCO

4) Three of the same letter, and a fourth distinct letter e.g. NANN

Obviously, none of these types overlap, and no other words exist outside of these types (you have no letters of multiplicity $4$, for example, so no words of one letter), so you just need to calculate how many exist for each type, which is much easier, and then add them together. Try and see if you can do this yourself.

EDIT: I should note that this will give you the same answer Kiran stated, $4436$, if that helps.


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