Probability of rearrangements of letters

[UniPhysics90]

If the letters of the word 'MINIMUM' are arranged in a line at random, what is the probability that the 3 M's are together at the beginning of the arrangement.

There are 2 methods outlined to solve this problem.

Method 1: (3/7)*(2/6)*(1/5)=(1/35) I understand this method completely.

The method I would like clarified is:

Method 2:

The statistical weight, w, w=7!/(3!2!1!1!)=420.
The number of ways of writing 'mmm****' where * is (i,n,u) is 4!/(2!1!1!)=12.
Thus p=12/420=1/35I get that the factorials are used as m is repeated 3 times, and i 2 times. I don't know why we divide by these, as in my mind there are 7! possible ways of arranging the letters, albeit some of them repeats. For probability, why is this used? Thanks

 

[awkward]

If the letters of the word 'MINIMUM' are arranged in a line at random, what is the probability that the 3 M's are together at the beginning of the arrangement.

There are 2 methods outlined to solve this problem.

Method 1: (3/7)*(2/6)*(1/5)=(1/35) I understand this method completely.

The method I would like clarified is:

Method 2:

The statistical weight, w, w=7!/(3!2!1!1!)=420.
The number of ways of writing 'mmm****' where * is (i,n,u) is 4!/(2!1!1!)=12.
Thus p=12/420=1/35


I get that the factorials are used as m is repeated 3 times, and i 2 times. I don't know why we divide by these, as in my mind there are 7! possible ways of arranging the letters, albeit some of them repeats. For probability, why is this used? Thanks

The reason for calculating w is to count the number of possible arrangements of the letters in MINIMUM, all of which we assume to be equally likely. If all the letters were distinct, the number of arrangements would be 7!. But there are duplicates: 3 Ms, 2 Is. So imagine we put tags on the Ms and Is and consider the set \{M_1, I_1, N, I_2, M_2, U, M_3\}. With the tags, there are now 7! arrangements. Now divide by 3! to compensate for the fact that the Ms are actually indistinguishable and divide by 2! to compensate for the fact that the Is as indistinguishable.

Does this make sense to you? It may help to know that the numbers you are calculating are multinomial coefficents: see

http://en.wikipedia.org/wiki/Multinomial_coefficient

 

[AlephZero]

Here are two slightly different versions of your method 2:

(a) Suppose the letters were written on 7 balls in a bag.
There are 7! possible ways to draw the balls from the bag (but not every way looks different, because there are 3 M's and 2 I's)

The number of ways to draw the 3 Ms, followed by the 4 other letters, is
3! 4!

So the probability is 3! 4! / 7! = 1/35

(b). Now, suppose you are writing the letters in a line.
The number of diifferent ways to write the 7 letters is 7! / 2! 3! (because there are 3 Ms and 2 Is)
The number of different ways to write 3 Ms followed by the other 4 letters is
1 (for the 3 Ms) x 4!/2! (for the other four letters, includng the two I's)

So the probabilty is
(4!/2!) / (7!/2!3!)
= 2! 3! 4! / 2! 7! = 1/35

Hope that helps.

 

[UniPhysics90]

The reason for calculating w is to count the number of possible arrangements of the letters in MINIMUM, all of which we assume to be equally likely. If all the letters were distinct, the number of arrangements would be 7!. But there are duplicates: 3 Ms, 2 Is. So imagine we put tags on the Ms and Is and consider the set \{M_1, I_1, N, I_2, M_2, U, M_3\}. With the tags, there are now 7! arrangements. Now divide by 3! to compensate for the fact that the Ms are actually indistinguishable and divide by 2! to compensate for the fact that the Is as indistinguishable.

Does this make sense to you? It may help to know that the numbers you are calculating are multinomial coefficents: see

http://en.wikipedia.org/wiki/Multinomial_coefficient

The idea makes sense, in terms of number of different rearrangements. But if there are multiple of the same letter, does this not increase the probability of getting that combination?

ie m(1),m(2),m(3)i(1)n(1)i(2)u(1) is the same arrangement in terms of just letters as m(2),m(1),m(3)i(1)n(1)i(2)u(1) etc.

 

[AlephZero]

But if there are multiple of the same letter, does this not increase the probability of getting that combination?

No, because you can divide all of the 7! possible arrangements into groups of 3! 2! arrangements where the Ms and Is are in the same position for all the arrangements in the group.

Each group contains the same number of arrangements, so you can either count the number of groups, or count the total number of arrangements. My previous post did it both ways.