Permutations: 4-letter words using letters from CALCULATE

how many ways can u spell a 4 letter word using letters from CALCULATE using:
1) all different letters
2) two "a"s and another identical letter pair
3) two identical letter pairs
4) one pair of identical leters and 2 other different letters

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So, this is what I have so far:

1) All different letters, so you have 6 (CALUTE) letters to choose from to make 4 letter scrambels. So, the answer is P(6,4) = 6!/2! = 360 possibilities

2) Two "a"s, and another identical pair. So, I assumed you could only have two other "l"s, or two other "c"s. So, using AACC, P(4;2,2) = 4!/2!2! = 6. Same thing with AALL. So, there are 12 possibilities

3) With any two identical pairs. So, same as 2), but you add another 6 possibilities in the case of CCLL. So, 18 possibilities.

4) I'm not sure how to tackle this. So far in class, we've only learned permutation formulas, nothing about combinations (the choose thing). So... is there a way to solve this using permutations only?

Thanks!

Re: Permutations

I will help you with #4.
There are three ‘doubles’ from which to select.
Once that is done, we have five letters left from which to select two.
Now we have four letters to arrange, but two are identical.
The number of ways this gives us is \(\displaystyle { 3 \choose 1}{ 5 \choose 2} \frac {4!}{2!}.\)

Re: Permutations

pka said:

I will help you with #4.
There are three ‘doubles’ from which to select.
Once that is done, we have five letters left from which to select two.
Now we have four letters to arrange, but two are identical.
The number of ways this gives us is \(\displaystyle { 3 \choose 1}{ 5 \choose 2} \frac {4!}{2!}.\)

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Thanks for your reply.
The thing is, I've only learned formulas involving permutations, not combinations. So, I haven't been introduced to the "choose" method. Is there a way to do this question, only dealing with factorials?

Re: Permutations

ty2391 said:

The thing is, I've only learned formulas involving permutations, not combinations. So, I haven't been introduced to the "choose" method. Is there a way to do this question, only dealing with factorials?

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Frankly, I find that just an unbelievable statement.
Having taught this area for years, I cannot understand your condition.
So I don’t think that I can help you. Sorry about that.
Maybe someone here can, I hope so.

Re: Permutations

I'm just taking a Senior high school data management course, and the first unit is Permutations and Organized Counting, while the second unit is Combinations and Binomial Theorem, and this question is in the first unit review. Thanks again though, I appreciate your help.

Re: Permutations

ty2391 said:

I'm just taking a Senior high school data management course, and the first unit is Permutations and Organized Counting, while the second unit is Combinations and Binomial Theorem, and this question is in the first unit review. Thanks again though, I appreciate your help.

Click to expand...

You can do this problem by simply counting.

There are three ways to select one of the double letters - that is pka's first term.

You have already chosen one double - you need to choose 2 letters from the remaining 5 ( 2 doubles and 3 singles).

you can think of it this way - you have five friends A, B, C, D & E - and you need to choose two of those people. They can be (AB), (AC), (AD), (AE), (BC), (BD), (BE), (CD), (CE) & (DE) - total 10 ways - and that is pka's second term.

Rest is permutation - which you know.