# How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga?

How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga?

I'm thinking we need permutations? I already tried finding each letter and still don't get it... any help?

So far I've broken it off to:-

$M=1$

$I=2$

$S=4$

$A=2$

$U=1$

$G=1$

However, I don't really know what to do next.

• Moo, I did. I tried doing that like how I would do it for "Mathematics". However Mississauga has like 4ms so it's tricky – PuppyLover101 Nov 7 '17 at 21:37
• First assume your word has only one of each letter. How many arrangements are there? Now consider repetitions. You have 4 s. These can be shuffled in $4!$ ways (which doesn't produce unique arrangements) so you need to divide by this number. Similarly for other repetitions. – LoMaPh Nov 7 '17 at 22:04
• Please edit your question to show your attempt. – N. F. Taussig Nov 8 '17 at 21:59

M=1

I=2

S=4

A=2

U=1

G=1

Case 1: All 4 letters are same

There is only one arrangement for this.

Case 2: 3 letters are similar

The three letters must be S. The remaining letter can be chosen in ${5 \choose 1}$ ways and these can be arranged in $\frac {4!}{3!}$ ways. This makes a total of 20 arrangements.

Case 3: 2 pairs are similar

The two repeating letters can be chosen in ${3 \choose 2}$ ways and can be arranged in $\frac {4!}{{2!}{2!}}$. This equals 18 arrangements.

Case 4: 2 are similar

The repeating letters can be chosen in ${3 \choose 1}$ ways and the remaining two in ${5 \choose 2}$ ways. These can be arranged in $\frac {4!}{2!}$ ways. This equals 360 arrangements.

Case 5: All 4 are different The 4 letters can be chosen in ${6 \choose 4}$ ways and can be arranged in $4!$ ways. This equals 360 ways.

Hence the total number of arrangements is $1+20+18+360+360=759$.

• amazing. But I have a question, why not have 6 cases for case 1? we can have MMMM IIII SSSS etc – PuppyLover101 Nov 9 '17 at 14:41
• You only have 1 of M, you cannot use it more than once. – user10984851 Nov 9 '17 at 15:05

Here's a hint as to one way to do it. Break it up into mutually-distinct cases:

• All four letters are different.
• One letter appears twice; two letters appear once.
• Two letters appear twice.
• One letter appears three times; one letter appears once.
• One letter appears four times.

Can you take it from here?

• Im really confused still... Like I tried that but that's the wrong answer still.. – PuppyLover101 Nov 8 '17 at 21:02
• Do you know the correct answer? If not, then tell me what you got for each case above. – John Nov 8 '17 at 21:19
• I don't.. Any idea if you can solve it completely? Like apparently this question is going to be on the test and I got "777 ways" as my answer but my teacher said I was close but still the wrong answer. – PuppyLover101 Nov 9 '17 at 6:28
• One of the other answerers (the one currently with the most votes) gave a complete, correct solution. (Well, it agrees with the result I got, so I think it's correct.) The rest of us were trying to give you a nudge in the right direction without showing you the whole solution, on the premise that you'll be stronger at the end of it having struggled with it a bit. When you ask future questions here I'd recommend showing your work a bit more so we can help where you're stuck; more learning will take place and all that ... – John Nov 9 '17 at 16:53

The word Mississauga has the following multiplicity: $M(1\times), I(2 \times ), S(4\times), A(2 \times), G(1\times), U( 1 \times)$

You want a word with $4$ letters, so lets take for example $(M,I,S,A,G,U)$. For a $4$-letter word you have a lot of combinations. Now you just have to write them down.

I'm going to do a few and let you do the rest.

For example if you have a $4$ letter word with one $M$ , one $I$, one $A$ and one $U$. You have $(1,1,0,1,0,1)$. For this you have $\frac{11!}{1!1!1!1!}$

If you take one $M$, two $I$ and one $S$ , you have $(1,2,1,0,0,0)$. For this you have $\frac{11!}{1!2!1!}$

You have to write all of combinations possible of $(M,I,S,A,G,U)$ with the sum value of $4$, where $M,I,S,A,G,U$ can vary up to their multiplicity respectively. In the end you sum up everything and get your result. However the word Missisauga has alot of combinations, so it might take some time.

• The book "Mathematics for Algorithm and Systems Analysis by Edward A. Bender & S. Gill Williamson" talks about this type of problem and it's free. – Numbermind Nov 8 '17 at 22:54

Considering letters in the order $M,I,S,A,G,U,$ the approach given by MathScientist can be encapsulated in the formula

Coefficient of $x^4$ in $4!(1+x)(1+x+\frac{x^2}{2!})(1+x+\frac{x^2}2 + \frac{x^3}{3!} +\frac{x^4}{4!})(1+x+\frac{x^2}{2!})(1+x)(1+x)$

which, of course, can be condensed and ordered systematically to

Coefficient of $x^4$ in $4!(1+x)^3(1+x + \frac{x^2}{2!})^2 (1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac {x^4}{4!})$