Question

the number of distinguishable permutations of: A,L,G,E,B,R,A is 2520, but how do you find that?

Answer 1

What did you try with this that didn't work? 7 factorial?

Answer 2

7C6 7P6 6!*7! 7!/(7!*6!)
I tried them all in confusion

Answer 3

The way I look at it is by looking at each place in the "word" as a numbered spot.

_ _ _ _ _ _ _
1 2 3 4 5 6 7

We start with seven letters. We have seven choices to put into the first spot. Then six choices to put into the second spot, because we placed one down for the first spot. Five choices for the third spot. Four for the fourth. And so on. We can multiply to take all the possible combinations of these 7 letters.
7*6*5*4*3*2*1 = Total # of permuations for 7 letters.

That would be great to understand. There is one subtle thing to address to get 2520 though.

Answer 4

there's two a's!

Answer 5

Exactly! So for all the combinations that we have: A_____A, we have a duplicate: A_____A (the A's are swapped around). We can't tell these apart, so they are indistinguishable!

So from 7!, we divide by the number of ways we can arrange the two A's. Which is just 2! = 2.
7! / 2 = 2520. Does that make more sense?

Answer 6

Yes, a lot more! Thank you :)

Answer 7

Glad to help! :)

If we had three A's instead of two, it is good not to forget that we want the combinations of arranging A 's and not just the quantity of them to divide off.
7! / 3! = 7! / 6, not 7! / 3.
And if we have more than one duplicate, we just divide the number of combinations of each.
Say, we had two A's and two G 's, 7! / 2! / 2! = 7! / 4.

Just some stuff that can come up to throw people off!

Answer 8

so for M,I,S,S,I,S,S,I,P,P,I would it be 11!/4!/4!/2! ?

Answer 9

or is it not factorial?

Answer 10

Four S's, four I's, and two P's. Then a lonely M.
Looks good!

Answer 11

11! / 4! / 4! / 2! is good. :)
The factorials all tell us the number of arrangements of (11 letters, 4 letters, and 2 letters). That is what we are interested in. Dividing off the 4! and 2! removes duplicate arrangements for us.

Answer 12

so, 34650? I didn't use parentheses

Answer 13

That appears correct.
Also, out of curiosity, the problem was stated very close the same to the first? I seem to remember some problems being phrased intentionally to trick people as well.

Answer 14

Yeah, sometimes there's a lot of methods to try and trick you