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Total number of arrangement of the word $”\bf{MATHEMATICS}”$ in which

no two same letter occur together.

$\bf{My\; Try::}$ Here word contain $\bf{2M,2A,2T,H,E,I,C,S}$

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So first we will arrange $\bf{H,E,I,C,S}$ as $\bf{5!}$ ways

$$\bf{-H-E-I-C-S-}$$

No we have $6$ gap and we can arrange $\bf{2M,2T,2A}$ as $\displaystyle \frac{6!}{2!\cdot 2!\cdot 2!}$

So total number of ways is $\displaystyle \bf{5!\times \frac{6!}{2!\cdot 2!\cdot 2!}}$

Is my solution is right, If not then how can we calculate it.

Help me

Thanks

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First, let’s count the number of distinguishable arrangements of the word MATHEMATICS, which has eleven letters. We can fill two of the eleven positions with an M in $\binom{11}{2}$ ways. We can fill two of the remaining nine positions with an A in $\binom{9}{2}$ ways. We can fill two of the remaining seven positions with a T in $\binom{7}{2}$ ways. The five remaining letters can be permuted in $5!$ ways. Hence, the number of arrangements of the letters of the word MATHEMATICS is

$$\binom{11}{2}\binom{9}{2}\binom{7}{2} \cdot 5! = \frac{11!}{9!2!} \cdot \frac{9!}{7!2!} \cdot \frac{7!}{5!2!} \cdot 5! = \frac{11!}{2!2!2!}$$

From these arrangements, we must exclude those in which two adjacent letters are the same.

We use the Inclusion-Exclusion Principle.

Consider those arrangements in which two adjacent letters are the same. Suppose, for example, that the two A’s are consecutive. Place them in a box. We now have ten objects to arrange, the other nine letters in the word MATHEMATICS and the box containing the two A’s. We can select two of the ten positions for the M’s in $\binom{10}{2}$ ways, two of the remaining eight positions for the T’s in $\binom{8}{2}$ ways, and arrange the other six objects in $6!$ ways. Thus, the number of arrangements in which the two A’s are adjacent is

$$\binom{10}{2}\binom{8}{2} \cdot 6! = \frac{10!}{2!2!}$$

An analogous argument applies to the two M’s and two T’s. Since there are three ways of choosing the pair of adjacent letters that are the same, the number of arrangements in which two consecutive letters are the same is

$$\binom{3}{1} \cdot \frac{10!}{2!2!}$$

Next, we count those arrangements in which two adjacent letters are the same and two other adjacent letters are the same. Suppose, for example, that the two A’s are adjacent and the two M’s are adjacent. Place the A’s in an amber box and the M’s in a maroon box. We now have nine objects to arrange, the two boxes, two T’s, and the other five letters. We can select two of the nine positions for the T’s in $\binom{9}{2}$ ways, then arrange the remaining seven objects in $7!$ ways. Hence, the number of arrangements in which the two A’s are adjacent and the two M’s are adjacent is

$$\binom{9}{2} \cdot 7! = \frac{9!}{2!}$$

Since there are $\binom{3}{2}$ ways to select two adjacent letters that are the same and two other adjacent letters that are the same, the number of arrangements of the word MATHEMATICS in which two adjacent letters are the same and two other adjacent letters are the same is

$$\binom{3}{2} \cdot \frac{9!}{2!}$$

Finally, we count those arrangements in which the two A’s are adjacent, the two M’s are adjacent, and the two T’s are adjacent. Place the A’s in an amber box, the M’s in a maroon box, and the T’s in a turquoise box. We then have eight objects to arrange, the three different color boxes and the five other letters. They can be arranged in $8!$ ways.

Thus, by the Inclusion-Exclusion Principle, the number of distinguishable arrangements of the word MATHEMATICS in which no two adjacent letters are the same is

$$\frac{11!}{2!2!2!} – \binom{3}{1} \cdot \frac{10!}{2!2!} + \binom{3}{2} \cdot \frac{9!}{2!} – 8!$$

My thanks to Barry Cipra for making me aware of the flaws in my first attempt to solve the problem.

Let’s lay out the problematic letters A, M, and T first and then make sure they get separated.

Without restriction, there are ${6\choose2,2,2}={6!\over2!2!2!}=90$ ways to arrange $2$ A’s, $2$ M’s, and $2$ T’s. Among these $6$ arrangements have all three letters doubled up (e.g., AAMMTT), $18$ have two letters doubled up (e.g., TAATMM), $36$ have one letter doubled up (e.g., TMAAMT), and $30$ have no letters doubled up (e.g., AMTMAT). Each of these counts requires a bit of thought, but I’ll take them as given.

We now need to wedge in the remaining five letters, C, E, H, I, and S, in a way that busts up any doubled-up letters. Let’s do one example carefully and then summarize the rest.

Suppose we have MMTTAA. Our first three wedge-in will bust up the pairs by attaching something immediately to the right of the first occurrence of each letter, i.e., (Mx)M(Ty)T(Az)A. This can be done in $5\times4\times3$ ways. The remaining two letters must then be wedged in among these $6$ groupings, and this can be done in $7\times8$ ways, for a total of $(5\times4\times3)\times(7\times8)$ arrangements.

If there are two doubled letters, the same approach gives of total of $(5\times4)\times(7\times8\times9)$ arrangements, and if there is only one doubled letter, it’s $5\times(7\times8\times9\times10)$. Finally, if there are no doubled letters, the remaining $5$ letters can be wedged in in ($7\times8\times9\times10\times11)$ ways.

To wrap things up, we just have to multiply these counts by their occurrences and add everything up:

$$6\times(5\times4\times3)\times(7\times8)+18\times(5\times4)\times(7\times8\times9)+36\times5\times(7\times8\times9\times10)\\+30\times(7\times8\times9\times10\times11)$$

If I’ve done all the arithmetic correctly, the final answer is $2{,}772{,}000$.

**Remark**: N. F. Taussig’s inclusion-exclusion approach leads to the same final answer, so I am somewhat more confident in my arithmetic. It’s a bit curious that $2{,}772{,}000=11\times10\times5\times7!$. In particular, the answer is divisible by $11$, the total number of letters, even though neither approach makes it obvious that that should be the case. So I wonder if that’s purely coincidental, or if there is some approach to the count that does make it obvious that the answer is divisible by the number of letters involved.

The word contains $\bf{2M,2A,2T,H,E,I,C,S}$

Use inclusion-exclusion to compute

[All ways] – [at least $1$ pair together] + [at least $2$ pairs together] – [all $3$ pairs together]

$= \dfrac{11!}{(2!)^3} – \dbinom31\dfrac{10!}{(2!^2} +\dbinom32\dfrac{9!}{2!} – {8!}$

$5!{6\choose {2,2,2}}=5!{6\choose 2}{4\choose 2}{2\choose 2}$

fill the gaps between letters $h,e,i,c,s$ $2$ at a time combined with all orderings of $h,e,i,c,s$. simplifies to same answer of $5!{{6!}\over{2!2!2!}}$

Here are two solutions generated using this tool of careerbless.com

[see generated question no.24 with the word ‘MATHEMATICS’)

Hint an alternative to check the correctness do parts it is safest .

\begin{align*}

\text{total arrangements} & – \text{ways where all letters M,A,T are together with the same letters}\\

& – \text{ways where M,T are together}\\

& – \text{ways where A,T are together}\\

& – \text{ways where A,M are together}\\

& – \text{ways where M are together}\\

& – \text{ways where A are together}\\

& – \text{ways where T are together}

\end{align*}

though its not healthy and competitive approach it ensures that all cases are taken.

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