| 5. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? | |
| A. 42000 | B. 48000 |
| C. 50400 | D. 47200 |
Answer: Option C
Explanation:
The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN[OOAIO].
Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and rest of the letters are different.
Number of ways to arrange these
letters
$=\dfrac{7!}{2!}=\dfrac{7×6×5×4×3×2×1}{2×1}=2520$
In the 5 vowels [OOAIO], 'O' occurs 3 and rest of the vowels are different.
Number of ways to arrange these vowels among themselves $=\dfrac{5!}{3!}=\dfrac{5×4×3×2×1}{3×2×1}=20$
Hence, required number of ways
$=2520×20=50400$
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- The correct option is A 50400In the word 'CORPORATION', we’ll treat the vowels OOAIO as a single letter. Thus, we have CRPRTN [OOAIO]. This has 7 [6 + 1] letters of which R occurs 2 times and the rest are different. Number of ways of arranging these letters =7!2!!=2520 Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged In 5!3!=20 ways Therefore, Required number of ways =[2520×20]=50400
- Question Detail
- 1. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours
- 2. Evaluate permutation equation \begin{aligned} ^{59}{P}_3 \end{aligned}
- 3. In how many words can be formed by using all letters of the word BHOPAL
- 4. Evaluate permutation equation \begin{aligned} ^{75}{P}_2\end{aligned}
- 5. How many words can be formed by using all letters of TIHAR
- How many different ways CORPORATION be arranged so that vowels come together?
- How many different ways can the letters of the word CORPORATION be arranged so that the vowels always come together Mcq?
- How many ways can the letters of the word CORPORATION be arranged so that vowels always occupy even places?
- How many words can be formed from the word CORPORATION?
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Solution
The correct option is A 50400In the word 'CORPORATION', we’ll treat the vowels OOAIO as a single letter. Thus, we have CRPRTN [OOAIO]. This has 7 [6 + 1] letters of which R occurs 2 times and the rest are different. Number of ways of arranging these letters =7!2!!=2520 Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged In 5!3!=20 ways Therefore, Required number of ways =[2520×20]=50400
Question Detail
- 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
Similar Questions :
1. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours
- 12
- 24
- 48
- 168
Answer: Option B
Explanation:
This question seems to be a bit typical, isn't, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24
Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND
1 Blue, so we multiplied.
2. 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}
3. 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}
4. Evaluate permutation equation \begin{aligned} ^{75}{P}_2\end{aligned}
- 5200
- 5300
- 5450
- 5550
Answer: Option D
Explanation:
\begin{aligned}
^n{P}_r = \frac{n!}{[n-r]!} \\
^{75}{P}_2 = \frac{75!}{[75-2]!}
\\
= \frac{75*74*73!}{[73]!} \\
= 5550
\end{aligned}
5. How many words can be formed by using all letters of TIHAR
- 100
- 120
- 140
- 160
Answer: Option B
Explanation:
First thing to understand in this question is that it is a permutation question.
Total number of words = 5
Required number =
\begin{aligned}
^5{P}_5 = 5! \\
= 5*4*3*2*1 = 120
\end{aligned}
Read more from - Permutation and Combination Questions Answers
Fernando Fidel Flores 8 years ago
what has 7 letters?
I only counted 6 consonants and 5 vowels.mastguru 8 years ago replied
Hello Fernando,
As per solution we have written like CRPRTN[OOAIO]
Here we take it 7 letters , 6 [CRPRTN] + 1 [OOAIO], as we need to vowel always come together, this way we counted them 7. Then solved the question.
Hope it helped you..Gabriela 9 years ago
THANK YOU SOOO MUCH, IT WAS FRO A HOMEWORK AND I WAS SO CONFUSED, I KNOW HOW TO DO COMBINATIONS AND PERMUTATIONS BUT I WASNT SURE OF WICH OF THEM TO USE AND HOW TO WRITE IT....THANKS :]
- 1
How many different ways CORPORATION be arranged so that vowels come together?
So, the total number of ways of arranging the letters of the word 'CORPORATION' be arranged so that the vowels always come together are 7!
How many different ways can the letters of the word CORPORATION be arranged so that the vowels always come together Mcq?
Therefore, Required number of ways =[2520×20]=50400.
How many ways can the letters of the word CORPORATION be arranged so that vowels always occupy even places?
60 ways. Originally Answered: In how many ways can the letters of word “corporation” be arranged so that each vowel occupies even places? It is an 11 lettered word with 5 vowels [3 o's and 1 a & 1 i] and 6 consonants. Moreover, there are 5 even places and 6 odd places.
How many words can be formed from the word CORPORATION?
There are in all 11 letters in the word `CORPORATION`. Since, Repetition is not allowed, there are 8 different letters that can be used to form 3-letter word. Therefore, total number of words that can be formed = 8P3 = [8 × 7 × 6] = 336.