# Math Help - Counting Identical Elements

1. ## Counting Identical Elements

How many arrangements of letters of word TRANSITION are possible if
the Is and Ns and Ts are together.

Here is my attempt:
Group Is, Ns, and Ts together, and arrange: 5! ways
Arrange amonst Is, Ns, Ts, removing repeition: 6! / 2!2!2!

This is not what the book has, which contains 5040 as answer.

2. Originally Posted by Lukybear
How many arrangements of letters of word TRANSITION are possible if
the Is and Ns and Ts are together.

Here is my attempt:
Group Is, Ns, and Ts together, and arrange: 5! ways
Arrange amonst Is, Ns, Ts, removing repeition: 6! / 2!2!2!

This is not what the book has, which contains 5040 as answer.
When you require that for example the T's be together, that means you treat it as a single entity, so since there are 7 distinct letters, the answer is simply 7! as the book says.

3. Hello, Lukybear!

undefined is absolutely correct.

$\text{How many arrangements of letters of word TRANSITION}$
$\text{ are possible if the I's and N's and T's are together?}$

Since the double-letters are to be adjacent, duct-tape them together.

Then we have 7 "letters" to arrange: . $\bigg \{A,\;O,\;R,\;S,\;\boxed{II},\;\boxed{NN},\;\boxed {TT}\bigg \}$

Therefore, there are: . $7! \,=\,5040$ arrangements.

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I see another interpretation of this problem.

It should have said: . $\begin{Bmatrix}\text{"The I's are adjacent,} \\ \text{and the N's are adjacent,} \\ \text{and the T's are adjacent."} \end{Bmatrix}$

Instead it said: "The I's and N's and T's are together."
. . which could mean that we have, for example, $NTINIT$
. . somewhere in the arrangement.

So we have 5 "letters" to arrange: . $\bigg\{A,\:O,\:R,\:S,\:\boxed{IINNTT} \bigg\}$
. . There are: . $5!= 120$ ways.

But $\boxed{IINNTT}$ can be arranged in: . $\displaystyle {6\choose2,2,2} = 90 \text{ ways.}$

Therefore, there are: . $120 \times 90 \:=\:10,800$ arrangements.