Question Detail
- 17280
- 4320
- 720
- 80
Answer: Option A
Explanation:
word SIGNATURE contains total 9 letters.
There are four vowels in this word, I, A, U and E
Make it as, SGNTR[IAUE], consider all vowels as 1 letter for now
So total letter are 6.
6
letters can be arranged in 6! ways = 720 ways
Vowels can be arranged in themselves in 4! ways = 24 ways
Required number of ways = 720*24 = 17280
Similar Questions :
1. Evaluate permutation equation
\begin{aligned} ^{59}{P}_3 \end{aligned}
- 195052
- 195053
- 195054
- 185054
Answer: Option C
Explanation:
\begin{aligned}
^n{P}_r = \frac{n!}{[n-r]!} \\
^{59}{P}_3 = \frac{59!}{[56]!} \\
= \frac{59 * 58 * 57 * 56!}{[56]!} \\
= 195054
\end{aligned}
2. In how many words can be formed by using all letters of the word BHOPAL
- 420
- 520
- 620
- 720
Answer: Option D
Explanation:
Required number
\begin{aligned}
= 6! \\
= 6*5*4*3*2*1 \\
= 720
\end{aligned}
3. How many words can be formed from the letters of the word "SIGNATURE" so that vowels always come together.
- 17280
- 4320
- 720
- 80
Answer: Option A
Explanation:
word SIGNATURE contains total 9 letters.
There are four vowels in this word, I, A, U and E
Make it as, SGNTR[IAUE], consider all vowels as 1 letter for now
So total letter are 6.
6 letters can be arranged in 6! ways = 720 ways
Vowels can be arranged in themselves in 4! ways = 24 ways
Required number of ways = 720*24 = 17280
4. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done
- 456
- 556
- 656
- 756
Answer: Option D
Explanation:
From
a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
[5 men] or [4 men and 1 woman] or [3 men and 2 woman]
\begin{aligned}
[^{5}{C}_{5}] + [^{5}{C}_{4} * ^{6}{C}_{1}] + \\
+ [^{5}{C}_{3} * ^{6}{C}_{2}] \\
= \left[\dfrac{7 \times 6 }{2 \times 1}\right] + \left[\left[ \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right] \times 6 \right] + \\ \left[\left[ \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right]
\times \left[ \dfrac{6 \times 5}{2 \times 1} \right] \right] \\
= 21 + 210 + 525 = 756
\end{aligned}
5. In how many ways can the letters of the word "CORPORATION" be arranged so that vowels always come together.
- 5760
- 50400
- 2880
- None of above
Answer: Option B
Explanation:
Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN[OOAIO]
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400
Read more from - Permutation and Combination Questions Answers
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
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.