How many ways can one rearrange the letters in the word ATTENTION such that the first letter is a T but the second one isn't?

Question

The number of arrangements of all letters in the word ATTENTION that can be arranged so that each arrangement begins with a T, and the second letter is not a T is

my work:

There are $9$ letters in the word ATTENTION so that means there are nine spaces so you would do $9!=9\times8\times7\times6\times5\times4\times3\times2\times1=362880$ of arrangements

Is this the correct way to do this problem? If not, how do you do it?

Answer 1

Approach via rule of product.

• Choose the three positions of the $T$'s
• Choose the two positions of the $N$'s from those remaining
• Arrange the remaining letters

Note that when choosing the three positions of the $T$'s, the first position must be included and the second position must not be included. So... this is effectively the same as choosing two positions from the last seven positions.

Note that regardless what choice was made in this first step, although the types of choices change for the later steps the number of choices is always the same.

$$\binom{7}{2}\times\binom{6}{2}\times 4!$$

Answer 2

Here is an alternative way to compute it

The first letter need to be a "T": $3$ possibilities

The second letter is not a "T": $6$ possibilities.

The we put the remaining letters, giving $$3\times6\times7\times6\times5\times4\times3\times2\times1$$ But this number consider the "T"s and the "N"s to be distincts. Since they are not, we need to divide this number by $3\times2\times1$ for the "T"s and $2\times1$ for the "N"s

Final answer: $$\frac{3\times6\times7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times2\times1}=7560\text{ possibilities}$$