# How many ways can you arrange the letters of the word “LETTER”? [duplicate]

I know that the answer is $\frac{6!}{2!2!}$, but I'm just curious why this is so. I got no intuition for this stuff.

## marked as duplicate by user147263, rogerl, Marconius, N. F. Taussig, JMPOct 31 '15 at 1:41

• Similar question: math.stackexchange.com/questions/236471/… – GrawCube Oct 30 '15 at 19:30
• Another way to see that is: You pick the letter L. You have to choose one place in _ _ _ _ _ _ to put the L, so 6 ways. Now pick the E's. You have 5 spaces to put those E's, that is, you have $\binom{5}{2}$ ways to put the two E's. Now the T's (do it yourself) and the R – Leafar Oct 30 '15 at 19:30
• Yeah, of course, a nice question. – Bhaskara-III Oct 30 '15 at 19:30
• @Leafar I like that way of doing it best. It makes the most sense to me. It just doesn't quite make sense how (6 choose 2)(4 choose 1)(3 choose 2)(1 choose 1) = 180 = (6 choose 1)(5 choose 1)(4 choose 2)(2 choose 2). Like order REALLY doesn't matter here, but that doesn't make any sense to me. Like, after you choose one, there will be less options to choose from, and so on and so on... – Bliebervik Oct 30 '15 at 19:41

Pick the two places where the $E$'s go: $_6C_2.$ Of the four remaining spaces, pick where the two $T$'s go: $_4C_2.$ Pick where the $L$ goes: $_2C_1.$ Pick where the $R$ goes ($_1C_1$; there's no choice).

This gives:

$$N = \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{1!1!} \cdot \frac{1!}{1!0!} = \frac{6!}{2!2!1!1!} = \frac{6!}{2!2!}.$$

The form just to the left of the answer is a multinomial form, and specifically counts unique permutations of objects, some of which are identical. You have a total of $6$ letters, $2$ of one kind, $2$ of another kind, $1$ of a third kind, and $1$ of a fourth kind.

Edit: As you noted, you get the same answer regardless of which order you place the letters. So, let's do $L,E,R,T$ in that order:

$$N = \frac{6!}{1!5!} \cdot \frac{5!}{2!3!} \cdot \frac{3!}{1!2!} \cdot \frac{2!}{2!0!} = \frac{6!}{2!2!}.$$

Notice that you can always cancel something in the denominator of some term with a term in the numerator immediately to the right (in this case, $5!, 3!, 2!$.) In the first case, it was $4!, 2!, 1!$.

This expresses mathematically something that makes sense: If you are counting arrangements of something, and doing it correctly, it shouldn't matter how you count them.

• But why can we do that? After we take the 2 E's, there will be 4 numbers left, etc. But we could also start with the one to get an answer of, say (6 choose 1)(5 choose 2)(3 choose 1)(2 choose 2), and this would also be correct. Why is that? – Bliebervik Oct 30 '15 at 19:43
• Edited my answer. – John Oct 30 '15 at 20:25
• I like you. Thanks! – Bliebervik Oct 30 '15 at 20:32

You have got $6$ letters in the word "LETTER". Now keeping the other letters unchanged, if you change the positions of the $2$ $T$'s or the $2$ $E$'s only, you will get no new arrangement.

Hence, you first consider the arrangement of all $6$ letters together considering all of them as distinct i.e. $6!$. Then you divide the answer by $2!$ twice, once for the $2$ $T$'s and then the $2$ $E$'s due to the above logic.

You have 6 Letters in total.
For the 1 position you have 6 possible letters.
If you choose one, you have 5 letter left for the second position.
If you choose one, you have 4 letter left for the third position.
etc. until you have 1 letter for the last position.
=> to distribute all letters you have 6! possibilities.

Now since you have duplicate letters, switsching the two T's or E's makes no difference. And we have to remove the duplicate answers from our solution. The number of possible permutations for 2 Ts is 2!, same for the Es.