How many ways can the letters of the word Learn be arranged so that the vowels always come together?

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History of Indian Constitution

15 Questions 15 Marks 9 Mins

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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In how many different ways can the letters of the word "CORPORATION" be arranged so that all the vowels always come together?

1. 810
2. 1440
3. 2880
4. 50400

Answer (Detailed Solution Below)

Option 4 : 50400

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NIMCET 2020 Official Paper

120 Questions 480 Marks 120 Mins

Concept:

• The number of ways in which r objects can be arranged in n places is nPr = \(\rm \dfrac{n!}{(n - r)!}\).
• The number of ways in which all n objects can be arranged among themselves = nPn = n!.
• The number of ways in which n objects, out of which p, q, r etc. are of same type, can be arranged is: \(\rm \dfrac{n!}{p!\ q!\ r!\ ...}\).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

Calculation:

The word CORPORATION has 11 letters out of which 6 are consonants (CRPRTN) and 5 are vowels (OOAIO).

Considering the objects of the same type, the number of arrangements of these vowels will be \(\rm \dfrac{5!}{3!}\) = 20.

Since, the vowels have to be together, we can say that we have to arrange the groups (C), (R), (P), (R), (T), (N) and (OOAIO) among themselves.

Considering the objects of the same type, this can be done in \(\rm \dfrac{7!}{2!}\) = 2520 ways.

And, total number of arrangements of all the letters = [Number of arrangements of (C), (R), (P), (R), (T), (N) and (OOAIO)] × [Number of arrangements of (OOAIO)] = 20 × 2520 = 50400.

Example from the textbook:

Question:

Sorry people I know most of u are too smart for this question, but I really need help so for questions c, d, e and f I have solved already, but I don't think my method of solving them is correct. G is the one I struggled for hours.

info I already know 8 letters, 3 vowels (A,E,I) and 5 consonants (C,T,R,N,G)

My Working Out

for c, i did 3 x 4 x 5 x 4 x 3 which equals to 720 which is correct.

for d, i did (3 x 4) - (2 x 3) first which equals to 6 I then multiply it by 3, 2, 5, 4 together therefore the final answer is 720. See in the first part I multiply 3 (the number of vowels) by 4 the number of position as any of the three vowels can fit in any of the 4 spaces. The questions asks for two vowels, since 1 vowel is already in one space the second vowel has two of the 3 vowels to take from and any of the two vowels can fit into the 3 remaining spaces. I subtract, because my intuition tells me to do so.

for e, I did (3 x 4) - (2 x 3) - (1 x 2) first which equals to 4 I then multiply it by 3, 2, 1, 5 together, making the answer is 120 which is the right answer. Here, I follow the my own principle from d. So I'm not sure if I'm right.

for f, I got the total arrangements from question a minus the total amount of words without vowels. To get the words without vowels i found all four letter arrangement words with consonants only which is 120. 1680 (from a) - 120 = 1560 (Correct)

for g, I totally don't understand it, but the answer is 18 000.

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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