What ways the letters of the word cricket can be arranged to form the different new words so that the vowels always come together?

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

Answer

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.

Overview

This tool lists out all the arrangements possible using letters of a word under various conditions. This can be used to verify answers of the questions related to calculation of the number of arrangements using letters of a word.

This tool programmatically generates all the arrangements possible. If you want to find out the number of arrangements mathematically, use Permutations Calculator

For example, consider the following question.

How many words with or without meaning can be formed using the letters of 'CRICKET' such that all the vowels must come together?

The answer of the above problem is $720$. Using this tool, it is possible to generate all these $720$ arrangements programmatically.

At the same time, Permutations Calculator can be used for a mathematical solution to this problem as provided below.

The word 'CRICKET' has $7$ letters where $2$ are vowels [I, E].

Vowels must come together. Therefore, group these vowels and consider it as a single letter.
i.e., CRCKT, [IE]

Thus we have total $6$ letters where C occurs $2$ times.

Number of ways to arrange these $6$ letters
$=6!2!=360$

All the $2$ vowels are different.
Number of ways to arrange these $2$ vowels among themselves
$=2!=2$

Required number of ways
$=360×2=720$

Note: This tool uses JavaScript for generating the number of permutations and can be slow for large strings.

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Aptitude Permutation And CombinationPage 2

Answer is: CGiven words contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
then, the letters to be arranged are DGHTR[AUE]
These 6 letters can be arranged in 6P6 = 6! = 720 ways
The vowels in the group [AUE] may be arranged in 3! = 6 ways
required number of words = [720 x 6] = 4320.

Answer is: CRequired number of ways = 15C11 = 15C[15 - 11] = 15C4
Required number of ways = [15 x 14 x 13 x 12]/[4 x 3 x 2 x 1] = 1365

Answer is: CRequired number of ways = [8C5 x 10C6]
Required number of ways = [8C3 x 10C4]
Required number of ways = [[8 x 7 x 6]/[3 x 2 x 1]] x [[10 x 9 x 8 x 7]/[4 x 3 x 2 x 1]]
Required number of ways = 11760

Answer is: CWe may have[1 black and 2 non-black] or [2 black and 1 non-black] or [3 black].
Required number of ways = [3C1 x 6C2] + [3C2 x 6C1] + [3C3]
Required number of ways = [3 x [6 x 5]/[2 x 1]] + [[3 x 2]/[2 x 1] x 6] + 1
Required number of ways = 45 + 18 + 1 = 64

Answer is: BThere are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
[1] [2] [3] [4] [5] [6]
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = [6 x 6] = 36.

Answer is: ARequired number of ways = 7C5 x 3C2 = 7C2 x 3C1 = [[7 x 6]/[2 x 1] x 3] = 63.

Answer is: CLOGARITHMS contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
Required number of words = 10P4
Required number of words = [10 x 9 x 8 x 7]
Required number of words = 5040.

Answer is: CIn the word MATHEMATICS, we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS [AEAI].
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/[2!][2!] = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12.
Required number of words = [10080 x 12] = 120960.

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