 # Five letter words are formed each containing two consonants and three vowels out of the letter of the word EQUATION. In how many of these are the two consonants always together? Verified
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Hint: This type of problem is solved by using the formula of permutation and combination. First take out total consonant and vowel given in word then make the combination with the required consonants and vowels by using the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where n is the total number of terms and r is the chosen terms.

The given letter is EQUATION.
There are three consonants (Q,T,N) and five vowels (E,U,A,I,O) in the letter EQUATION.
So, the total number of constants is 3 and the total number of vowels is 5.
Now we make to make the combination by using the two consonants and three vowels then,
Total number of combination of choosing 2 consonants out of 3 consonants is $^3{C_2}$
So we get, the total combination of constants as,
$^3{C_2} = \dfrac{{3!}}{{2!1!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} = 3$
Total number of combination of choosing 3 vowels out of 5 vowels is $^5{C_3}$
So we get, the total combination of vowels as,
$^5{C_3} = \dfrac{{5!}}{{3!2!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2 \times 1}} = 10$
There are fives letters in the word then the number of combinations of these fives words are $5! = 120$
So total number of combination is
$3 \times 10 \times 120 = 3600$
But we have to find the combination only when two consonants come together, so instead of taking a combination of all five letters we carry out a combination of 4 letters and consider the constants as one letter. So the number of ways these four letters arrange is $4! = 24$
Now, a condition arises that we cannot separate the consonants but these constants can also change their position at one place. So the number of combinations these consonants arrange is $2! = 2$
Hence the total number of combination of making a five letter word when two consonants are together is: $3 \times 10 \times 24 \times 2 = 1440$
Hence the required number of combinations is 1440.
So, the correct answer is “1400”.

Note: We can also solve this problem by using the permutation instead of combination by using the formula of $^n{P_r} = \dfrac{{n!}}{{r!}}$.
Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.