1

The word MATHEMATIK (in german) has 10 letters.

I want to know how many different words can be created by using the letters of this word. However, the same letter can never be next to each other.

So for example, MAATHEMTIK is not allowed.

I did the following:

$n = 10$

$k_1 = 2$ (M)

$k_2 = 2$ (A)

$k_3 = 2$ (T)

$k_4 = 1$ (H)

$k_5 = 1$ (E)

$k_6 = 1$ (I)

$k_7 = 1$ (K)

I use the formula $\frac{n!}{k_1! \cdot k_2! ... \cdot k_n! }$ and get:

$$\frac{13!}{2! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1! \cdot 1!} = 778377600$$

This is the number of all possible words (permutations). Now we need to subtract the number of words which can be created by having the same letter next to each other, i.e.

  • MAATHEMTIK
  • AAMTHEMTIK
  • MTAAHEMTIK
  • etc.

But I don't know how to figure that out.

stepbysteptomathpro
  • 685

3 Answers3

2

Please apply Principle of Inclusion Exclusion. It is a more standard approach.

Your answer would simply be

$\displaystyle \frac{10!}{2! \ 2! \ 2!} - \binom{3}{1} \ \frac{9!}{2! \ 2!} + \binom{3}{2} \ \frac{8!}{2!} - \binom{3}{3} \ 7! = 236880$

Some details -

The first term is unrestricted number of arrangements. Now we subtract from it all arrangements where either of $AA, MM, TT$ are together. If you take $AA$ together as one letter, you now have $9$ letters and the number of arrangements are $\frac{9!}{2! \ 2!}$. Now you have same number of arrangements for $MM$ and $TT$. But these overcount arrangements where two identical letters (say AA, MM) are together. As we subtracted them, we need to add them back. The same follows for the next.

Math Lover
  • 51,819
1

First of all the letters are :
2×A,2×M,2×T,H,E,I,K.
Now we will do as the following first of all it is clear that there is $7!$ different ways of making a word just by using the letters A,M,T,H,E,I,K.
note that there are $8$ different places for putting a letter in this word (one before the first letter one between the first and second letters one between the second and third and so on) so if we want to place a second letter A there are $8-2$ different possible places since there are two places adjacent to the first A and we should not place our second A in those places. So there is $7!×6$ different methods of putting 2×A,M,T,H,E,I,K such that no two same letters are adjacent.
Now in the same way there are 9 different places for putting the second M but two of them are out of access since the second M should not be adjacent to the first letter M. So there is $7!×6×7$ different methods of putting 2×A,2×M,T,H,E,I,K such that no two same letters are adjacent.
And finally for putting the second T there are 10 different places two of which or not allowed to have another T in them. So at last there are $7!×6×7×8$ different words having the same letters as MATEMATIKS such that no two same letters are adjacent.
Now note that in this counting, each valid word will be counted exactly 8 times and this is the reason:
If for a 10-letter word $W$ made of the letters of MATEMATIKS, we choose one A, one M and one T and the H,E,I,K as the initial letters, then by removing the other 3 letters there will be 7 initial letters, using which, by going through the foresaid procedure $W$ can be built. So technically each word is produced 8 times each time by a different initial 7-letter. (That's clearly because there are 2×2×2 choices for choosing one A, one M and one T to be the initial ones out of two A, two M and two T. Also note that because $W$ is a word in which no two same letters are adjacent, all of these initial 7-letters that produce $W$ are pairwisely different)
Hence any word is counted 8 times and the answer is $7!×6×7×8÷8=7!×6×7$

Aryan
  • 1,508
  • 1
    You will get duplicates. You need more rigorous counting or apply principle of inclusion exclusion. For example, take two arrangements of your first $7$ letters that you are permuting - M A T H E I K and A T H M E I K. Now take the case when you add A, M, T in empty places one by one of the first arrangement and make A T M A T H M E I K. Now again for the second, you can add A T M in empty places one by one and make A T M A T H M E I K. – Math Lover Jan 10 '21 at 21:09
  • @MathLover So 1693440 is false too? – stepbysteptomathpro Jan 10 '21 at 21:14
  • Math Lover. Thanks I'll correct it right away. The whole counting is true yet there is a small error – Aryan Jan 10 '21 at 21:21
  • Now you have some missing cases. Please see my answer. – Math Lover Jan 10 '21 at 21:38
  • What answer is missing. Can you be more clear? – Aryan Jan 10 '21 at 21:39
  • When you simply divide your answer by $8$, you have some arrangements missing. You should check. I have not gotten into details this time. But your answer does not match the answer using P.I.E. – Math Lover Jan 10 '21 at 21:42
  • I checked it there's nothing missing when you divide it by 8. And it's because two same initial 7-letters can't produce same word and so all of the initial 7-letters of a word are different and there are are exactly 8 initial 7-letters for each final word. I admit PIE is an easier approach but for the version of the problem that has variable number of letters it will just crash – Aryan Jan 10 '21 at 21:50
  • The method you are using is an important method and I do use it. But in cases like this, it takes a lot of careful counting. I can give you some direction on where you may have missing cases - when you arrange letters like this, you will never take the second $A$ to be next to first $A$. So you will miss cases like $A M A$. You were getting those cases from other $A M T$ but when you divide by $8$, the counting goes wrong as they were never overcounted. – Math Lover Jan 10 '21 at 22:11
  • No thats not the case. In my counting that'll be like {something}AM{something} and you want to put another A right after M, but that's one of the valid places for the second A in my counting so I do count it – Aryan Jan 10 '21 at 22:23
  • That's what I said. You do count it but do not overcount. So they need to be taken out before dividing by 8 and then added back. Anyway I have an extended discussions warning but if you decide to find out, I will move this to chat and we can discuss there. – Math Lover Jan 11 '21 at 03:50
  • Let us continue this discussion in chat. – Aryan Jan 11 '21 at 05:47
0

Someone mentioned another solution involving the permutations of "Mississippi," but it makes things unnecessarily complicated.

Simple solution:

Assume initially that all the letters of 'mathematik' are distinct, subscripting them if necessary. So the first $m$ is labelled $m_1$ and the second is $m_2$. Then with 10 "distinct" letters (not really distinct, just different b/c of the subscripts) there are $10!$ permutations.

Now to deal with the overcounting due to the subscripts: for each subscripted letter divide by the factorial of its count. For example, since $m$ appears twice, divide by 2!. Likewise divide by 2! because of the 'a' appearing twice, and again by 2! because of the 't' appearing twice.

So the final answer is ${10! \over {2! 2!2!}}$.

In general, if a letter appears 3 times, divide by 3! to compensate for overcounting, etc.

[edit]: I forgot to incorporate the fact that a letter can't appear next to the same letter. Please see Aryan's answer below.

Hank Igoe
  • 1,408
  • Thank you! I was about to apply the "principle" the person used in the other answer, but it looked extremely complicated – stepbysteptomathpro Jan 10 '21 at 20:09
  • 1
    You're welcome, it's my pleasure. Math is hard enough even without adding extra complications. btw, may I ask how you got your avatar name? – Hank Igoe Jan 10 '21 at 20:11
  • 1
    You mean the "Retarded Programmer"? It took me forever to learn programming (Java), so I just consider myself a retard when it comes to computer science – stepbysteptomathpro Jan 10 '21 at 20:12
  • You're solution is definitely wrong. The number you presented is the number of all words with the letters of MATEMATIKS not the ones that don't have same letters adjacent to each other. – Aryan Jan 10 '21 at 20:20