Total number of arrangement of the word $"\bf{MATHEMATICS}"$ in which
no two same letter occur together.
$\bf{My\; Try::}$ Here word contain $\bf{2M,2A,2T,H,E,I,C,S}$
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
Best Answer
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.