The company, hewerrtt, would like to make letter codes using all of the letters in the word hewerrtt. how many codes can be made

Question

ira [324]

3 years ago

5

The company, hewerrtt, would like to make letter codes using all of the letters in the word hewerrtt. how many codes can be made

from all the letters in this word?

Mathematics

2 answers:

shtirl [24] · Answer 1 · 2022-03-07T01:00:42+03:00

<span>If the company, hewerrtt, would like to make letter codes using all of the letters in the word hewerrtt, they would be able to make five codes. Altogether, there are eight letters in the word, but there are duplicates of the letter e, the letter r, and the letter t, so that takes out three possibilities. Three subtracted from eight is equal to five, therefore the company can make five codes using the company name.</span>

Gemiola [76] · Answer 2 · 2022-03-07T02:08:34+03:00

The number of codes that can be made from all the letters in word hewerrtt is $\boxed{\bf 5040}$ .

Further explanation:

It is given that the company hewerrtt would like to make letter codes using all of the letters in the word hewerrtt.

We have to find the permutation of the word hewerrtt with repeating letters that is $e$ , $r$ and $t$ .

Consider the total number of permutation as $P$ .

The formula that is used to calculate the permutation of a word for repeating letters is given as follows,

$\fbox{\begin\\\ \math P=\dfrac{n!}{r!\cdot s!\cdot t!\cdot \cdot}\\\end{minispace}}$ ......(1)

Here $n$ represents total number of words and $r!\cdot s!\cdot t!\cdot \cdot \cdot$ represents the repeating letters of a given words.

The number of letters in the word, “hewerrtt” is $8$ and the number of repeating letters is $3$ that is $e,r$ and $t$ .

Here $e$ repeats two times, $r$ repeats two times and also $t$ repeats two times.

Therefore the equation (1) can be written as follows,

$\fbox{\begin\\\ \math P=\dfrac{n!}{r!\cdot s!\cdot t!}\\\end{minispace}}$

Substitute $8$ for $n$ , $2$ for $r$ , $2$ for $s$ and $2$ for $t$ in above equation.

$\begin{aligned}P&=\dfrac{8!}{2!\cdot 2!\cdot 2!}\\&=\dfrac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2!}{2\cdot 1\cdot 2\cdot 1\cdot 2!}\\&=\dfrac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3}{2\cdot 2}\\&=5040\end{aligned}$