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My attempt :-

I made all the possible cases by fixing the positions of O in word, for example :-

  1. O at the 1st place and 3rd place
  2. O at the 1st place and 4th place
  3. O at the 1st place and 5th place
  4. O at the 1st place and 6th place
  5. O at the 1st place and 7rd place
  6. O at the 1st place and 8th place
  7. O at the 2nd place and 4th place . . . . .

This process seems too ineffecient for solving this problem.

Official answer is given as $\frac{8!}{2!}$ /3 = $6720$

I understand that $\frac{8!}{2!}$ is the total 8 letter rearrangements of the word but why in the solution is the author dividing this by 3 ?

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    $\begingroup$ Once you arrange L,C,K,D,W in 8 places there is only one way to place O,N,O $\endgroup$
    – Hari Shankar
    Commented Jul 11, 2023 at 6:10
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    $\begingroup$ This one looks like it would be fun on the codegolf SE. $\endgroup$
    – David S
    Commented Jul 11, 2023 at 21:43
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    $\begingroup$ Divide by three because you want those with -O-N-O- but neither -O-O-N- nor -N-O-O- and each is equally common $\endgroup$
    – Henry
    Commented Jul 12, 2023 at 9:26
  • $\begingroup$ @David, yes by all means; I thought the same thing $\endgroup$
    – JosephDoggie
    Commented Jul 12, 2023 at 13:21

5 Answers 5

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Another way to do it is to first place $-O - N - O -$
and then place the remaining $5$ letters one by one.

The first letter, say $L,$ can be placed in $4$ ways, the next in $-O-L-N-O-\;\;5$ ways, and so on, thus simply

$4\cdot5\cdot6\cdot7\cdot8 = 6720$

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    $\begingroup$ Thanks alot @true blue anil for the wonderful answer :) , Could you also please address my doubt of why we are dividing by 3 at the end in the solution I mentioned in my post $\endgroup$
    – Vasu Gupta
    Commented Jul 12, 2023 at 12:34
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    $\begingroup$ @VasuGupta: 1. Glad to be of help ! 2. Forget about ONO and first place the other $5$ letters in $8$ spots, which can be done in $^8C_5\times 5! = 8!/3!$ ways. ONO can then be placed in only one way in the remaining spots $\endgroup$
    – true blue anil
    Commented Jul 12, 2023 at 12:54
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    $\begingroup$ @VasuGupta: 1. The permutation symbol is not much favoured on this forum, else I could directly have said: Permute $5$ letters other than ONO in $8$ slots in $^8P_5 = 8!/3!$ ways. ONO , of course, can then be placed in the remaining slots in only one way. 2 Glad that you liked my answer ! $\;\;$ :) $\endgroup$
    – true blue anil
    Commented Jul 12, 2023 at 13:12
  • $\begingroup$ WOW. That's a huge number. $\endgroup$
    – Vanity Slug - codidact.com
    Commented Jul 12, 2023 at 16:45
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There are $\frac{8!}{2}$ different words possible. A word has the letters (o,o,n) in one of these orders $\{o,o,n\}$, $\{n,o,o\}$ or $\{o,n,o\}.$ These sets are in bijection (switch one of the o’s to n in your bijection) so they all have the same number of elements. Therefore answer is $$\frac{\frac{8!}{2}}{3}.$$

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    $\begingroup$ Hi @Artlind, thank you for the answer... I am not aware of what bijections are , could you please elaborate what you meant to show by the fact that (o,o,n) will appear only in 3 orders as mentioned $\endgroup$
    – Vasu Gupta
    Commented Jul 12, 2023 at 12:32
  • $\begingroup$ Hi @VasuGupta, for example in lOckdOwN, letters (o,o,n) appear in this order (o,o,n). In the workd dlONckwO they are like that (o,n,o). Bijection is a function that completly maps two ensembles meaning you can associate each element of the first ensemble to a unique element in the second.. Use it to show that ensembles have the same size. For us here, lets consider these 2 ensembles: -E1 : words where letters (o,o,n) appear in this order -E2 : words where letters (o,o,n) appear in the order (o,n,o) You can create a bijection from E1 to E2 by changing the second o to a n. $\endgroup$
    – Artlind
    Commented Jul 12, 2023 at 22:19
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You know that the Os and N have to go in the order O…N…O. So the arrangements you want correspond to placing the other letters LCKDW (which are all distinct), together with three blanks where the O,N,O will go — i.e. arrangements of LCKDW•••, where e.g. L•CKD•W• corresponds to LOCKDNWO.

So you just want the number of arrangements of LCKDW••• — 8 objects, 3 identical and the rest distinct — which is $8!/3!$.

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    $\begingroup$ Hi Peter, thank you for the answer , I am not able to understand how ONO are taken as 3 identical objects , could you elaborate on it iplease $\endgroup$
    – Vasu Gupta
    Commented Jul 12, 2023 at 12:37
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    $\begingroup$ Since the order of the O,N,O is forced, you just need “blanks” to mark which three places they’ll go in. The arrangement of the other letters plus the three blanks completely determines the overall arrangement, because there’s exactly one way to fill in the blanks with O,N,O. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jul 12, 2023 at 12:45
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Define an equivalence relation $\sim$ on these words such that $u\sim v$ if the set of positions of letters O,O,N is the same in $u$ and $v$ (for instance, this set is $\{2,6,8\}$ in LOCKDOWN).

Now it is easy to check that each equivalence class is of size $3$ since there are $3$ possible positions for N, but you want to keep only one of these possibilities, so the final answer is divided by $3$.

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Start off with eight blank spaces to place the eight letters in.

Initial blanks: _ _ _ _ _ _ _ _

Now place five of these letters in spaces: specifically, choose a spot for all the letters except for O, N, and O

Example: D L _ C W _ K _

Now, how many ways were there to do this? There were 8 choices for the L, then once that was placed 7 choices for the C, 6 for the K and so forth.

So far, we could arrange these five letters $8*7*6*5*4$ possible ways.

But here's the neat part: those three remaining blanks already have their letters fixed! After all, since the N is always between two Os, the leftmost blank must be O, then the middle blank is N, then the rightmost blank the other O. This is true no matter where these final three blanks end up.

So the total number of ways we could arrange the letters with the N between the Os is exactly the number of ways we could arrange the other five letters: namely $8!/3!$

= $8*7*6*5*4$

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  • $\begingroup$ NOTE: I consider this mostly just another version of @PeterLeFanuLumsdaine 's excellent answer. I included it because I thought the details might help someone who thought about this differently, $\endgroup$
    – Gandalfmeansme
    Commented Jul 12, 2023 at 17:37

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