If the different permutations of the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E appears?
Answer
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Hint: The letters are arranged in a dictionary in alphabetical order. Therefore, in this question, we have to find the number of words starting with A, B, C, D to get the number of words before the first word starting with E appears.
Complete step-by-step answer:
We first try to find the total number of words starting with A. For this, we fix A at the beginning and find the different permutation of the other letters after that.
As the total number of letters in EXAMINATION is 11, the words starting with A should be of the form $\text{A }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }$. Now, after taking out A, the remaining letters are E, X, M, I, N, A, T, I, O, N.
We know that the number of distinct permutations of n objects (having k distinct objects repeated one or more times) in which object 1 is repeated $p_1$ times, object 2 is repeated $p_2$ times etc. is given by $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!{{p}_{3}}!...{{p}_{k}}!}\text{ }....................\text{(1}\text{.1)}$
However, we note that of these remaining letters, there are two I’s and two N’s which are identical. Thus, using equation (1.1) the number of distinct permutation of the remaining letters are $\dfrac{10!}{2!2!}=907200\text{ }..............\text{(1}\text{.2)}$.
Thus, the number of words from the letters of EXAMINATION are 907200.
As B, C and D are not present in the letters of the word EXAMINATION, the total number of words in the list before the first word starting with E is equal to the number of words starting with A, which is 907200.
Note: In this question it is necessary to divide $10!$ by $2!$ and $2!$ in equation (1.2) because the words in which the two I’s or two N’s would be interchanged would correspond to the same word in the dictionary.
Complete step-by-step answer:
We first try to find the total number of words starting with A. For this, we fix A at the beginning and find the different permutation of the other letters after that.
As the total number of letters in EXAMINATION is 11, the words starting with A should be of the form $\text{A }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }$. Now, after taking out A, the remaining letters are E, X, M, I, N, A, T, I, O, N.
We know that the number of distinct permutations of n objects (having k distinct objects repeated one or more times) in which object 1 is repeated $p_1$ times, object 2 is repeated $p_2$ times etc. is given by $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!{{p}_{3}}!...{{p}_{k}}!}\text{ }....................\text{(1}\text{.1)}$
However, we note that of these remaining letters, there are two I’s and two N’s which are identical. Thus, using equation (1.1) the number of distinct permutation of the remaining letters are $\dfrac{10!}{2!2!}=907200\text{ }..............\text{(1}\text{.2)}$.
Thus, the number of words from the letters of EXAMINATION are 907200.
As B, C and D are not present in the letters of the word EXAMINATION, the total number of words in the list before the first word starting with E is equal to the number of words starting with A, which is 907200.
Note: In this question it is necessary to divide $10!$ by $2!$ and $2!$ in equation (1.2) because the words in which the two I’s or two N’s would be interchanged would correspond to the same word in the dictionary.
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