Permutations/Combinations - Could someone please check my work?

In summary, the conversation was about a student seeking help with their assignment on Permutations and Combinations. They discussed three out of five questions, with the first question being answered correctly and the second and third being discussed and revised. The correct answer for the third question was determined to be 176 ways.
  • #1
rocketboy
243
1
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.
 
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  • #2
I don't need to know the exact answers, just if I'm correct or not.

Thanks.
 
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  • #3
You got #1 right. It is indeed (11!)/(4!*4!*2!) = 34,650.
--------------------------------------------------------

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. o:)
 
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  • #4
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. o:)
 
  • #5
IB said:
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. o:)

Is this correct? Can anybody verify please?
 
  • #6
IB said:
Okay, for #3, there can be many cases (possibilities-which letters we choose, not their arrangements in the 4 letter word):

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

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

Good, I had that.

IB said:
(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

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.

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

Alright, I had this.

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

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?
 
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1. How do you calculate permutations and combinations?

Permutations and combinations are calculated using different formulas. For permutations, the formula is n! / (n-r)! where n is the total number of items and r is the number of items being selected. For combinations, the formula is n! / r!(n-r)! where n is the total number of items and r is the number of items being selected.

2. What is the difference between permutations and combinations?

The main difference between permutations and combinations is the order in which the items are selected. Permutations take into account the order of selected items, while combinations do not. For example, selecting three letters from the alphabet in the order ABC is a permutation, while selecting the same three letters in any order is a combination.

3. Can you give an example of permutations and combinations?

An example of permutations is the number of ways 5 people can be arranged in a line, which is 5! = 120. An example of combinations is the number of ways 3 letters can be selected from the alphabet, which is 26! / (3!(26-3)!) = 260.

4. When should I use permutations and when should I use combinations?

Permutations should be used when the order of the selected items is important, such as arranging people in a line or selecting a password with specific characters in a specific order. Combinations should be used when the order of the selected items is not important, such as selecting a committee or choosing toppings for a pizza.

5. Could you check my work for permutations and combinations?

Yes, I can check your work for permutations and combinations. Make sure to double-check your calculations and use the correct formula for each situation. If you are still unsure, you can provide your work and I can review it for you.

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