Math Help - Permutation

1. Permutation

I have spent all evening trying to solve the following by many different ways to no avail. Would you please show me how to modify the ordinary formula for combination and permutation to suit the question:

Find the number of (a) combinations and (b) permutations of four letters each that can be made from the letters of the word Tennessee?

2. Originally Posted by novice
Find the number of (a) combinations and (b) permutations of four letters each that can be made from the letters of the word Tennessee?
.
I assume that you want four “letter words” made from the letters in Tennessee.
There are five cases: I. no letter repeated; II. one letter repeated; III. two letter repeated;
IV. three letters repeated; and V. four letters repeated.
$\begin{array}{lll}
\text{Case} & \text{Combin} & \text{Permut} \\
I & 1 & {4!} \\
{II} & {(3)(3)} & \displaystyle{\frac{{4!}}
{{2!}}} \\
{III} & {(3)(2)} & \displaystyle{\frac{{4!}}
{{(2!)^2 }}} \\
{IV} & 3 & \displaystyle{\frac{{4!}}
{{3!}}} \\
V & 1 & 1 \\ \end{array}$

3. Originally Posted by Plato
.
I assume that you want four “letter words” made from the letters in Tennessee.
There are five cases: I. no letter repeated; II. one letter repeated; III. two letter repeated;
IV. three letters repeated; and V. four letters repeated.
$\begin{array}{lll}
\text{Case} & \text{Combin} & \text{Permut} \\
I & 1 & {4!} \\
{II} & {(3)(3)} & \displaystyle{\frac{{4!}}
{{2!}}} \\
{III} & {(3)(2)} & \displaystyle{\frac{{4!}}
{{(2!)^2 }}} \\
{IV} & 3 & \displaystyle{\frac{{4!}}
{{3!}}} \\
V & 1 & 1 \\ \end{array}$
Plato, thank you for answering my question. I typed the question verbatim. I think it meant 4-letter words; other wise it will be too easy.

Could you explain why we need not consider the fact that we must use 9 letters to form 4-letter words, i.e.

In Case 1 for instance: n=9, k=4 (for 4-letter words), and r=4 (repeated letters) so that

9P4= n!/[(n-k)!r!] <--I know this is wrong but don't know why.

Pardon me for not knowing how to use latex for the factorials.
I appreciate the time given me for the help.

4. Hello everyone
Originally Posted by Plato
.
I assume that you want four “letter words” made from the letters in Tennessee.
There are five cases: I. no letter repeated; II. one letter repeated; III. two letter repeated;
IV. three letters repeated; and V. four letters repeated.
$\begin{array}{lll}
\text{Case} & \text{Combin} & \text{Permut} \\
I & 1 & {4!} \\
{II} & {(3)(3)} & \displaystyle{\frac{{4!}}
{{2!}}} \\
{III} & {(3)(2)} & \displaystyle{\frac{{4!}}
{{(2!)^2 }}} \\
{IV} & 3 & \displaystyle{\frac{{4!}}
{{3!}}} \\
V & 1 & 1 \\ \end{array}$
May I amplify Plato's solution a little, and query one of the results.

Case I. This is where we select 1 of each of the 4 different letters; so that's ${^4C_4}=1$ selection and $4!=24$ arrangements of these 4 different letters.

Case II. One letter repeated; for example: e, e, n, s.

There are 3 ways of selecting which letter is to be repeated, and then ${^3C_2} = 3$ ways of selecting which two of the remaining 3 different letters will accompany the repeated letter. So that's $3\times 3 =9$ selections.

We can then arrange 4 letters which contain one pair of repeated letters in $\frac{4!}{2!}=12$ ways.

Case III. Two letters repeated: e, e, n, n, for instance. I think that's just ${^3C_2} = 3$ ways of selecting which 2 of the 3 repeated letters we'll use. So I get a different answer here.

Then I agree with the answer of $\frac{4!}{(2!)^2}=6$ arrangements of 4 letters with 2 repeated pairs.

Case IV. I think this means 1 letter repeated 3 times. So that's the letter e with one of the remaining 3. That's 3 choices and $\frac{4!}{3!}=4$ arrangements of 4 items with a group of 3 'like' items.

Case V. This means 1 letter repeated 4 times. So that's 1 selection (all e's) and 1 arrangement.

So, when we add up, we get 17 selections (combinations) and 47 arrangements (permutations).

5. Grandad, You are correct.
Someday I will learn not to copy and paste so much.
Thanks

6. Originally Posted by Grandad
Hello everyoneMay I amplify Plato's solution a little, and query one of the results.

Case I. This is where we select 1 of each of the 4 different letters; so that's ${^4C_4}=1$ selection and $4!=24$ arrangements of these 4 different letters.

Case II. One letter repeated; for example: e, e, n, s.

There are 3 ways of selecting which letter is to be repeated, and then ${^3C_2} = 3$ ways of selecting which two of the remaining 3 different letters will accompany the repeated letter. So that's $3\times 3 =9$ selections.

We can then arrange 4 letters which contain one pair of repeated letters in $\frac{4!}{2!}=12$ ways.

Case III. Two letters repeated: e, e, n, n, for instance. I think that's just ${^3C_2} = 3$ ways of selecting which 2 of the 3 repeated letters we'll use. So I get a different answer here.

Then I agree with the answer of $\frac{4!}{(2!)^2}=6$ arrangements of 4 letters with 2 repeated pairs.

Case IV. I think this means 1 letter repeated 3 times. So that's the letter e with one of the remaining 3. That's 3 choices and $\frac{4!}{3!}=4$ arrangements of 4 items with a group of 3 'like' items.

Case V. This means 1 letter repeated 4 times. So that's 1 selection (all e's) and 1 arrangement.

So, when we add up, we get 17 selections (combinations) and 47 arrangements (permutations).

Originally Posted by Plato
Grandad, You are correct.
Someday I will learn not to copy and paste so much.
Thanks

Plato,

I have more questions on the permutations for case II, III, and IV.
They seem to imply one choice of repeated items.

I am thinking:

Case II: $\frac{4!}{2!}\times 3$ since there are three choices for the doublets.

Case III: $\frac{4!}{(2!)^2}\times 3!$ since there are 3! ways of pairing the doublets.

Case IV: $\frac{4!}{3!}\times 3$ since there are 3 other distinct characters that can accompany the triplets.

I am still not too sure that I have included all scenarios.

Note: I am grateful that you and GRANDAD help me change my way of thinking. I could not make a headway last evening because I was too fixated by the rule which said An arrangment of any r $\leq$ n of objects, $P(n,r)=\frac{n!}{(n-r)!}$.

7. The answers we gave are correct.

The additions you made above are not correct.

8. Originally Posted by Plato
The answers we gave are correct.

The additions you made above are not correct.
Plato,

Thank you for reviewing my offer. You got me thinking again.
I listed the the following into the computer and do it un-mathematically.

I fed the each listed as follows and got 12 arrangement each for Case II:

12 for 2n's + s + e
12 for 2n's + s + t
12 for 2n's + t + e

12 for 2s' + n + e
12 for 2's + n + t
12 for 2's + e + t

12 for 2e's + t + s
12 for 2e's + t + n
12 for 2e's + s + n

Total arrangements for Case II = 108 arrangements

For Case III:

6 for 2e's + 2s's
6 for 2e's + 2t's
6 for 2s's + 2t's

Total 18 arrangements for Case III.

For Case IV:

4 for 3'e + t
4 for 3'e + s
4 for 3'e + n

Total 12 arrangements.

For total inclusive arrangements is 163.

I see your work everyday. You guys are very sharp.

9. Originally Posted by novice
For total inclusive arrangements is 163.
Good gravey there are some hard-to-convince people.
One more time; the last time.
There are only 47 different 'four-letter words' using the leters in TENNESSEE.

10. Pardon me Plato. I have so much to learn from you than you from me. We correlated our solution at school. None of us have the same number. We went crazy.