In how many ways can we arrange the word coffeehouse so that all the vowels come together?

Answer

Verified

Hint: Here, we are required to arrange the letters in the given word ‘FACTOR’. Thus, we will use Permutations to ‘arrange’ the letters keeping in mind that all the letters in the given word are unique. Thus, applying the formula and solving the factorial, we will be able to find the required ways of arrangement of letters of the given word.

Formula Used:
We will use the following formulas:
1. ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ , where $n$ is the total number of letters and $r$ represents the number of letters to be arranged.
2. $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.

Complete step-by-step answer:
In order to find the arrangement of the word ‘FACTOR’,
First of all, we will observe that all the letters in this given word are unique and no word is the same or duplicate. Also, the number of letters in the word ‘FACTOR’ is 6.
Therefore, we will use Permutations to ‘arrange’ the 6 letters of the given word.
Thus, the formula is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Where, $n$ is the total number of letters and $r$ represents the number of letters to be arranged, i.e. $6$ in each case.
Thus, we get,
${}^6{P_6} = \dfrac{{6!}}{{\left( {6 - 6} \right)!}} = \dfrac{{6!}}{{0!}} = 6!$
Because, $0! = 1$
Now, the formula of expanding factorial is $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.
Hence, we get,
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 30 \times 24 = 720$

Therefore, we can arrange the letters in the word ‘FACTOR’ in 720 ways.
Thus, this is the required answer.

Note:
While solving this question, we should know the difference between permutations and combinations. Permutation is an act of arranging the numbers whereas combination is a method of selecting a group of numbers or elements in any order. Hence, Permutations and Combinations play a vital role to solve these types of questions. . Also, in order to answer this question, we should know that when we open a factorial then, we write it in the form of: $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$ as by factorial we mean that it a product of all the positive integers which are less than or equal to the given number but not less than 1.

In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together?A. 1440B. 120C. 720D. 360

Answer

Verified

Hint: To solve this problem we have to know about the concept of permutations and combinations. But here a simple concept is used. In any given word, the number of ways we can arrange the word by jumbling the letters is the number of letters present in the word factorial. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$ \Rightarrow n! = n(n - 1)(n - 2).......1$

Complete step by step answer:
Given the word TRAINER, we have to arrange the letters of the word in such a way that all the vowels in the word TRAINER should be together.
The number of vowels in the word TRAINER are = 3 vowels.
The three vowels in the word TRAINER are A, I, and E.
Now these three vowels should always be together and these vowels can be in any order, but they should be together.
Here the three vowels AIE can be arranged in 3 factorial ways, as there are 3 vowels, as given below:
The number of ways the 3 vowels AIE can be arranged is = $3!$
Now arranging the consonants other than the vowels is given by:
As the left out letters in the word TRAINER are TRNR.
The total no. of consonants left out are = 4 consonants.
Now these 4 consonants can be arranged in the following way:
As in the 4 letters TRNR, the letter R is repeated for 2 times, hence the letters TRNR can be arranged in :
$ \Rightarrow \dfrac{{4!}}{{2!}}$
But the letters TRNR are arranged along with the vowels A,I,E, which should be together always but in any order.
Hence we consider the three vowels as a single letter, now TRNR along with AIE can be arranged in:
$ \Rightarrow \dfrac{{5!}}{{2!}}$
But here the vowels can be arranged in $3!$ as already discussed before.
Thus the word TRAINER can be arranged so that the vowels always come together are given below:
$ \Rightarrow \dfrac{{5!}}{{2!}} \times 3! = \dfrac{{120 \times 6}}{2}$
$ \Rightarrow 360$

The number of ways the word TRAINER can be arranged so that the vowels always come together are 360.

Note: Here while solving such kind of problems if there is any word of $n$ letters and a letter is repeating for $r$ times in it, then it can be arranged in $\dfrac{{n!}}{{r!}}$ number of ways. If there are many letters repeating for a distinct number of times, such as a word of $n$ letters and ${r_1}$ repeated items, ${r_2}$ repeated items,…….${r_k}$ repeated items, then it is arranged in $\dfrac{{n!}}{{{r_1}!{r_2}!......{r_k}!}}$ number of ways.

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