# Permutations/Combinations - Could someone please check my work?

1. Nov 24, 2005

### rocketboy

Hey everyone,

I have an assignment on Permutations and Combinations and was wondering if some of you could go over my work and ensure that I have this aced. I really need to get my math mark up for University admissions so I need to ace this assignment.

Thanks a lot for your help!
-Jonathan

1) In how many distinguishable ways can the letters of the word MISSISSIPPI be arranged?

Number of distinguishable arrangements = 11!/(4!4!2!) = 34,650

2) Find the number of ways of arranging all the letters AAABBBCCDD in a row so that no two A's are side-by-side.

Number of arrangements = 7!/(3!3!2!2!) = 35

what I did: I paired the letter A up with another letter, by viewing them as 1 letter no 2 A's would ever be side by side. This makes 7 letters, and then I divided out the repeats.

3) How many distinguishable four letter arrangements can be made from teh letters of the word MISSISSIPPI?

Number of four letter arrangements = (ways with no repeats) + (ways with all 4 letters repeated) + (ways with 2 doubles) + (ways with a triple and a single)

= (4c4)x4 + (2c1) + (3c2)x(4!/2!2!) + (2c1)(1c1)x(4!/3!)
= 24 + 2 + 18 + 8
= 52

*****If it is unclear what I have done hear please ask me*****

Those are the first 3 questions out of 5. I want to make sure I'm on the right track before I finish the next 2, as they are significantly more difficult.

2. Nov 24, 2005

### rocketboy

I don't need to know the exact answers, just if i'm correct or not.

Thanks.

Last edited: Nov 24, 2005
3. Nov 26, 2005

### IB

You got #1 right. It is indeed (11!)/(4!*4!*2!) = 34,650.
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For #2:

Step one: calculate all possible ways taking into account repetitions: 10!/(3!*3!*2!*2!)

Step two: consider AAA as one letter and thus calculate in how many varinats they are to gether (8!/(3!*2!*2!)

Step three: consider AA as one letter and calculate in how many cases they are together: (9!/(3!2!2!)

Step four: substract step 3 and step 2 from step one
-----------------------------------------------------

For #3 I'm still thinking. The only solution I can see right now is to count it manually. But meh, that isn't a good approach to combinatorics. Allow me a few more minutes to think of a better solution. Will post again soon.

Last edited: Nov 26, 2005
4. Nov 26, 2005

### IB

Okay, for #3, there can be many cases (possibilities-which letters we choose, not their arrangements in the 4 letter word):

(a) All four letters are different (MISP) 4*3*2*1=24

(b) If two letters repeat (2I, 2P or 2S) AND there is M

There are 6 possibilities (M2IP/S; M2PI/S; M2SP/I).
Therefore: 6*([4*3*2*1]/2)=72

(c) If 2 and 2 letters repeat then there are 3 possibilities (2I2P; 2I2S; 2P2S)
Therefore: 3*([4*3*2*1]/4)=18

(d) If 3 letters repeat there are 3 possibilities (3SM; 3SP; 3SI)
Therefore 4*3=12

Therefore we obtain: 24+72+18+12=126

I hope you got the method.

5. Nov 28, 2005

### rocketboy

Is this correct? Can anybody verify please?

6. Nov 28, 2005

### rocketboy

Ok, that is what I was doing in my first post, so far so good.:tongue2:

Ooooh...kk i forgot this. THANKS IB!

EDIT: I just worked this part out myself, and got a different answer for this case.

There are 3 sets of letters to choose doubles from, and we want 1 of them to be chosen. Then we want 2 singles from the 3 sets of singles that are left to be chosen from. (ex, if 2 I's are chosen, we need 2 singles, fro meither S, P or M.) Then we have to multiply by 4! (for different positions) and divide by 2! (the repeats).

Thus we get (3c1)*(3c2)*(4!/2!) = 108 ways.

Hmmm....I don't think this is correct. I had the same idea, and now that I think of it I think we are both wrong.

for each of the triples that can occur, there has to be another letter present. This means that if there are 3 I's, then the fourth can be an S, M or P. And, they can be in four different spots. I's and S's are the only letters that can give tripples. For each of them, there is a fourth letter needed, one of 3 letters possible. Then, for each of these arrangements there are 4! ways of arranging the letters. Thus I think the answer for this case is (2c1)*(3c1)*4!

Where (2c1) means "two choose one" (combinations).

So there should be 144 in this case....wierd, that can't be.

Can anyone help?

EDIT: I forgot to divide that answer by 3! to get rid of the repeats. This means that there are 24 ways for this case.

IB, I think you also forgot to include the arrangements where there are only repeats, such as (IIII) or (SSSS). Thanks for your help though, I appreciate it. Without it I would never have seen my errors.

Well, I redid it, and I get a total of 176 ways.

Who is right? or are we both wrong?

Last edited: Nov 28, 2005