In how many ways the word grapes can be arranged so that vowels are always together

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Electric charges and coulomb's law (Basic)

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The formula to find a number of ways word can be arranged, so that the vowels do not come together is:

Total word(factorial) - (Total word - 1)factorial × (number of vowels)factorial

Given:

Total words = 5

Total word - 1 = 4

Number of vowels = 2

Substituting in the formula:

5! - (4! × 2!) = 120 - 48

= 72

Hence, 72 ways are there to arrange SCALE so that vowels do not come together.

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• Aptitude
• Permutation and combination

Correct Answer:

Description for Correct answer:
The word SACRED consists of 4 consonants [SCRD] and two vowels (AE). On keeping vowels together we get SCRD(AE).

Therefore, Number of arrangements = \( \Large 5! \times 2! \)

= \( \Large 5 \times 4 \times 3 \times 2 \times 1 \times 1 \times 2 \)

= 240

Part of solved Permutation and combination questions and answers : >> Aptitude >> Permutation and combination

Answer

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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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