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How many words with or without meaning can be formed using all the letters of the word "EQUATION" at a time if vowels and consonants occur together. My answer is $5!3!2!=1440$. Am I right?
asked Sep 12, 2015 at 14:47
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In the word "EQUATION" there are 3 consonants [Q,T,N] so there are $3!$ ways to arrange and 5 vowels [E,U,A,I,O] so $5!$ ways to arrange. In whole for both first consonant and then vowels or vice-versa, there are $2!$ ways . so in total $2! \cdot 5! \cdot 3!=1440$ ways.
answered Jun 15, 2017 at 4:39
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Answer
Verified
Hint – Pick out the vowels and consonants separately from the given word ‘COURAGE’ and use the combination formula for selecting r dissimilar things out of total n things
Now in the given word ‘COURAGE’ there are 3 consonants that are [c, r, g] and there are 4 vowels that are [o, u, a, e].
So out of 3 consonants a single consonant can be chosen in $^3{C_1}$ways.
Now \[^n{C_r} = \dfrac{{n!}}{{r![n - r]!}}\] using this we can solve $^3{C_1}$ which will be \[\dfrac{{3!}}{{1![3 -
1]!}} = \dfrac{{3!}}{{2!}} = \dfrac{{3 \times 2!}}{{2!}} = 3\]
So out of 4 vowels a single vowel can be chosen in $^4{C_1}$ways.
Again using the similar concept as above we can solve for $^4{C_1}$ which will be \[\dfrac{{4!}}{{1![4 - 1]!}} = \dfrac{{4!}}{{3!}} = \dfrac{{4 \times 3!}}{{3!}} = 4\]
Hence the total number of ways of choosing a vowel and a consonant will be $4 \times 3 = 12$ ways
Note- In English alphabet set we have in total 27 alphabets which can broadly be classified into two categories that are vowels and consonants. [A, E, I, O, U] are the vowels while rest are marked as consonants.
ADDITIONAL QUESTION BANK Q.58
Q.144] In how many ways can a consonant and a vowel be chosen out of the letters of the word `LOGARITHM’?
Correct
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The word `LOGARITHM’ consist of 9 letters,in which there are 6 consonantes and 3 vowels. The selection of a consonant from 6 consonants in done in 6C1 = 6 ways. while selection of a vowel out 3 vowels is done by 3 different way.
The number of selection of a vowel and consonant
= 6 × 3
= 18.
Answer : [a]
The letters of the word LOGARITHM are arranged at random. Find the probability that start with vowel and end with ends with consonant.
Solution
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways.
∴ n[S] = 9!
Let E be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in
the word LOGARITHM.
∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in 7P7 = 7! ways
∴ n[E] = 3 × 7! × 6
∴ P[E] = `["n"["E"]]/["n"["S"]`
= `[3 xx 7! xx 6]/[9!]`
Concept: Elementary Properties of Probability
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