Answer
Hint: First of all, we have a total of 8 places. Fill the first and the last places by 3 consonants in 3 x 2 ways. Now fill the remaining places with the remaining 6 letters in 6! ways. Remember that there is no constraint in filling places from second to seventh place.
Here, we are given a word EQUATION and we have to find the number of words formed by letters of word EQUATION such that the words end and begin with consonants.
First of all, we know that in English language, we have
a total of 5 vowels and these vowels are A, E, I, O, U.
Now, the vowels present in the given word EQUATION are E, U, A, I, O. Therefore, the number of vowels present in the word EQUATION = 5.
Also, we know that in English language, all the letters except vowels are known as consonants. We have a total 21 consonants in English.
Now, the consonants present in the given word EQUATION are Q, T, N.
Therefore, the number of consonants present in the word EQUATION = 3.
Therefore, our
word has \[\text{ }\]
Now, we have to form 8 letter words such that letters at the beginning and end are consonants.
\[\underline{3}\times \underline{6}\times \underline{5}\times \underline{4}\times \underline{3}\times \underline{2}\times \underline{1}\times \underline{2}\]
We know that we have to fill the first place with consonants only and we have a total of 3 consonants, so we can fill the first place in 3 ways. Also, we have to fill the eighth or the last place with consonants
only and we have the remaining 2 consonants, so we can fill the last place in 2 ways.
Now, there are constraints to fill the remaining 6 places with 6 letters. So, any one of the 6 letters can occupy second place. So, there are a total 6 ways to occupy second place.
Likewise, third place can be filled by any of the remaining 5 letters. So, there are a total of 5 ways to occupy third place.
Similarly, fourth place can be filled by any of the remaining 4 letters. So there are a total of
4 ways to occupy fourth place.
In the similar way, fifth, sixth and seventh places can be occupied by remaining 3, 2 and 1 letters respectively. So, there are 3, 2, and 1 ways to fill fifth, sixth and seventh places respectively.
Therefore, we get the total number of words ending and beginning with consonants
\[\begin{align}
& =3\times 6!\times 2 \\
& =3\times 6\times 5\times 4\times 3\times 2\times 1\times 2 \\
& =4320 \\
\end{align}\]
Hence,
we can form a total 4320 different words ending and beginning with consonants with the letters of the word EQUATION.
Note: Here, students get confused if repetition is allowed or not. As it is clearly mentioned that the first and last letters of the word are consonants and words should be formed by letters of the word EQUATION then consider that repetition is not allowed. Also, it would not make sense if repetition is allowed as we can make infinitely many words if the repetition is allowed. Also, some students try to do this question manually by writing different words and counting them one by one which is a very lengthy approach and can give wrong results.
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Answer
VerifiedHint: Permutations are the different ways in which a collection of items can be arranged. For example:The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC. Note that ABC and CBA are not the same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.The number of ways in which n things can be arranged, taken all at a time, \[{}^n{P_n}{\text{ }} = {\text{ }}n!\], called ‘n factorial.’
Complete step-by-step answer:
Total number of letters in “EQUATION” = 8.There are 5 vowels: a, e, i, o, u and 3 consonants : q, t, n.Since all the vowels and consonants have to occur together, both [AEIOU] and [QTN] can be assumed as single objects.Then they form 2 groups V[vowels] and C [consonants]We first arrange the 2 groups.The permutations of these 2 objects taken all at a time are counted: ${}^2{P_2} = 2! = 2$waysCorresponding to each of these permutations, Now the group V has 5 elements, they can be arranged in $5! = 120$ ways.Now the group C has 3 elements, they can be arranged in $3! = 6$ ways.Hence by multiplication principle, required number of words = $2! \times 5! \times 3!$the total no of ways = $1440$Therefore, 1440 words with or without meaning, can be formed using all the letters of the word ‘EQUATION’, at a time so that the vowels and consonants occur together.
Note: Always keep an eye on the keywords used in the question. The keywords can help you get the answer easily.
The keywords like-selection, choose, pick, and combination-indicates that it is a combination question.Keywords like-arrangement, ordered, unique- indicates that it is a permutation question.If keywords are not given, then visualize the scenario presented in the question and then think in terms of combination and arrangement.Last updated at May 29, 2018 by Teachoo
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Ex 7.3, 8 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter Exactly once? Number of letters in word EQUATION` = 8 n = 8 If all letters of the word used at a time r = 8 Different numbers formed = nPr = 8P8 = 8!/[8 8]! = 8!/0! = 8!/1 = 8! = 8 7 6 5 4 3 2 1 = 40320
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Catherine T.
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How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?