The number of ways the letters of the word MATHEMATICS could be arranged into a row would be?

Question

The number of ways the letters of the word MATHEMATICS could be arranged into a row would be?

I have look at a similar question to this but what I don't understand is that the solution says that "there are two Ms, two As, etc." and hence the answer is $ 11! /8! $. Would someone care to explain this for me? (I do understand how permutations work).

Answer 1

I think the answer is wrong (given I correctly understood the question).

This is a case of Permutation of multiset and the formula is simply ${11!\over2!2!2!} = {11!\over8}$

Answer 2

The number of permutations of $M_1 A_1 T_1HEM_2A_2T_2ICS$ is 11!.

Now divide this number by $(2!)^3=8$ (we have two copies of three letters). Hence the final answer is $$\frac{11!}{8}$$ (the $8$ without the factorial).

Answer 3

Histogram of MATHEMATICS:

• M occurs $2$ times
• A occurs $2$ times
• T occurs $2$ times
• H occurs $1$ time
• E occurs $1$ time
• I occurs $1$ time
• C occurs $1$ time
• S occurs $1$ time

Hence the number of arrangements is:

$$\frac{(2+2+2+1+1+1+1+1)!}{2!\times2!\times2!\times1!\times1!\times1!\times1!\times1!}$$

Answer 4

Imagine instead of having indistinguishable Ms, As and Ts, the letters are instead distinct, so you'd have M1, M2, A1, etc.
The number of permutations of the word is then just $11!$. Now, you decide to drop this distinction between M1 and M2 (and the As and the Ts).
For an arbitrary permutation, there's now $8 = 2 \cdot 2 \cdot 2$ permutations that look the same: the ones you get from swapping the M1 and M2, A1 and A2, T1 and T2. So your $11!$ is $8$ times the number of permutations of the word MATHEMATICS.

For a similar example: the number of permutations of BANANA would be $6!$ if you'd have distinguinshable As and Ns, but then if you'd permute A1, A2 and A3 in any way (and there's $3!$ such ways) and then dropped the distinction, the word would look the same. Applying a similar reasoning for the Ns, the total number of permutations would be $\frac{6!}{3! 2! 1!} = 60$.