[SOLVED] Counting with identical items

**lindah** · January 24th 2010, 03:17 AM

Hi all,
I am having problems with the following part of the question:

The Letters of TRANSITION can be arranged to give different combinations of the word

a) count the amount of combinations where the N is only the first letter and not the last

- I am thinking that the final letter has 6 possible outcomes
- the remaining 8 slots are accounted for - but my sticking point is if "I" or "T" is in the last spot - how to eliminate over-counting them

b) N is neither the first or last letter

Thank you in advance for any feedback

**Plato** · January 24th 2010, 04:32 AM

Originally Posted by lindah

The Letters of TRANSITION can be arranged to give different combinations of the word
a) count the amount of combinations where the N is only the first letter and not the last

I will show you a trick with subscripts: $T_1RAN_1SI_1T_2I_2ON_2$ .
Now we have ten different letters which can be arrange in $10!$ ways.
To remove the subscripts we divide. So there are $\frac{10!}{(2!)^3}$ ways to arrange TRANSITION.

To answer part a, we can forget the subscripts on the N's.
Put one of those in first place. Now we can put any of the eight non-N's in the last place and arrange the other eight between them.
We get $(8)\left(\frac{8!}{(2!)^2}\right)$ .

**awkward** · January 24th 2010, 04:42 AM

Originally Posted by lindah

Hi all,
I am having problems with the following part of the question:

The Letters of TRANSITION can be arranged to give different combinations of the word

a) count the amount of combinations where the N is only the first letter and not the last

- I am thinking that the final letter has 6 possible outcomes
- the remaining 8 slots are accounted for - but my sticking point is if "I" or "T" is in the last spot - how to eliminate over-counting them

b) N is neither the first or last letter

Thank you in advance for any feedback

Hi Lindah,

Here is an approach which should work.

We know an N will be in the first slot, so count the number of arrangements of TRANSITIO which will go in the next 9 positions. If we disregard the duplicated T and I, there would be 9! possibilities. But we have over-counted by a factor of 2! for the T and 2! for the I, so there are
$\frac{9!}{2! \; 2!}$ distinct arrangements.

Some of these arrangements end in N. So see if you can find the number which end in N and subtract from the total to find the answer to a).

Edit: Beaten to the punch by Plato! But we gave different solutions anyway.

**lindah** · January 24th 2010, 01:18 PM

Thank you very much Plato and awkward!

Originally Posted by awkward

Some of these arrangements end in N. So see if you can find the number which end in N and subtract from the total to find the answer to a).

Oh yes, so this would be $\frac{8!}{2! \; 2!}$ arrangements subtracted from $\frac{9!}{2! \; 2!}$ . ty

b) N is neither the first or last letter

I have misquoted this

- "the letter N is not at either end"

I currently do not possess the answer to this question, however my thinking is as follows:

There are 8 letters that can occupy the first position
This leaves 7 letters to fill the last position
The remaining 8 spots would be arranged by $\frac{8!}{2! \; 2! \; 2!}$ to account for over-counting "T", "I" & "N"

Would this be correct
$(7)(8)(\frac{8!}{2! \; 2! \; 2!})$ ?

tia

**lindah** · January 24th 2010, 01:46 PM

I have an additional problem w.r.t to counting:

"B is about to hang his 8 shirts in a wardrobe. He has four different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no 2 identical shirts are next to one another"

I've reasoned that the combinations will need to be divided by $2!\; 2! \;2! \;2!$ to account for the 4 identical styles.

I've tried Plato suggestions with subscripts naming the styles $A_1, A_2, B_1, B_2, C_1, C_2, D_1, D_2$ and tried to count a method as follows:

8 possible choices for the first position
6 choices for the second [excl. the identical]
5 choices for the third
After this I obtain varying counts such as 8x6x5x5x3x2x2x1
Am I on the right track?

Thank you in advance for any feedback

**Plato** · January 24th 2010, 01:54 PM

$\frac{8!}{(2!)^4}$

**lindah** · January 24th 2010, 02:01 PM

Originally Posted by Plato

$\frac{8!}{(2!)^4}$

Hi Plato,
The answer in the solutions is: 864
I initially thought of the above answer - but the identical shirts cannot be lined together would make it different?

**Plato** · January 24th 2010, 02:50 PM

I do apologize. I read that too quickly. I missed that no two.
The answer given is correct. But the method is difficult and hard to explain.
It involves repeated use of inclusion/exclusion.
$\sum\limits_{k = 0}^4 {\left( { - 1} \right)^k \binom{4}{k} \frac{{\left( {8 - k}\right)!}}{{2^{4 - k} }}} = 864$ .

That idea is rather advanced for a beginner at this.
It is a variant of the old no two partners together in a queue.

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