What is the number of possible words that can be made using the word jasmine such that the vowels always come together?

Note: . Anagrams are meaningful words made after rearranging all the letters of the word.
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Note There are 3 vowel letters and 4 consonant letters in the word jasmine. J is 10th, A is 1st, S is 19th, M is 13th, I is 9th, N is 14th, E is 5th, Letter of Alphabet series.

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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.

The word games Words With Friends, 4pics1Word, Word Chums, and Jumble which is by far one of the most successful of the word games. Jumble was created in 1954 - below, you will find the most unscrambled letters for each descramble word game that others have solved or decoded to make the word jasmine.

Is jasmine a scrabble word or can you use jasmine in Words With Friends? The probability of getting this word in scrabble is 1 out of every 343098 games and in Words With Friends it's 1 out of every 453918 games. This 7 letter 16 point scrabble word can be rearranged 5,040 ways. What other words can be made with the letters a, e, i, j, m, n, and s? There's 3 with 9 letters or less with the letters a, e, i, j, m, n, and s. Here is a list of 3 to try to get you more points.

1) In what ways the letters of the word "RUMOUR" can be arranged?

1. 180
2. 150
3. 200
4. 230

Answer: D

Answer with the explanation:

The word RUMOUR consists of 6 words in which R and U are repeated twice.
Therefore, the required number of permutations =

Hence, 180 words can be formed by arranging the word RUMOUR.

2) In what ways the letters of the word "PUZZLE" can be arranged to form the different new words so that the vowels always come together?

1. 280
2. 450
3. 630
4. 120

Answer: D

Answer with the explanation:

The word PUZZLE has 6 different letters.

As per the question, the vowels should always come together.
Now, let the vowels UE as a single entity.
Therefore, the number of letters is 5 (PZZL = 4 + UE = 1)
Since the total number of letters = 4+1 = 5
So the arrangement would be in 5P5 =

Note: we know that 0! = 1

Now, the vowels UE can be arranged in 2 different ways, i.e., 2P2 = 2! = 2*1 = 2 ways

Hence, the new words, which can be formed after rearranging the letters = 120 *2 = 240

As we known z is occurring twice in the word ‘PUZZLE’ so we will divide the 240 by 2.

So, the no. of permutation will be = 240/2 = 120

3) In what ways can a group of 6 boys and 2 girls be made out of the total of 7 boys and 3 girls?

1. 50
2. 120
3. 21
4. 20

Answer: C

Answer with the explanation:

We know that nCr = nC(n-r)

The combination of 6 boys out of 7 and 2 girls out of 3 can be represented as 7C6 + 3C2
Therefore, the required number of ways = 7C6 * 3C2 = 7C(7-6) * 3C(3-2) =

Hence, in 21 ways the group of 6 boys and 2 girls can be made.

4) Out of a group of 7 boys and 6 girls, five boys are selected to form a team so that at least 3 boys are there on the team. In how many ways can it be done?

1. 645
2. 734
3. 756
4. 612

Answer: C

Answer with the explanation:

We may have 5 men only, 4 men and 1 woman, and 3 men and 2 women in the committee.

So, the combination will be

as we know that

nCr=

So, (7C3 * 6C2) + (7C4 * 6C1) + (7C5)
Or,

Or, 525 +210+21 = 756

So, there are 756 ways to form a committee.

5) A box contains 2 red balls, 3 black balls, and 4 white balls. Find the number of ways by which 3 balls can be drawn from the box in which at least 1 black ball should be present.

1. 64
2. 48
3. 32
4. 96

Answer: A

Answer with the explanation:

The possible combination could be (1 black ball and 2 non-black balls), (2 black balls and 1 non- black ball), and (only 3 black balls).

Therefore the required number of combinations = (3C1 * 6C2) + (3C2 * 6C1) + (3C3)
r,

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