Question

How many 6-letter words with distinct letters in each can be formed using the letters of the word EDUCATION? How many of these begin with I
(a) ${}^{9}{{P}_{6}},{}^{8}{{P}_{5}}$
(b) ${}^{9}{{P}_{6}},{}^{9}{{P}_{5}}$
(c) ${}^{8}{{P}_{6}},{}^{8}{{P}_{5}}$
(d) ${}^{8}{{P}_{6}},{}^{8}{{P}_{4}}$

Answer 1

Hint: Here, we should know the proper meaning of permutation i.e. given as the various ways in which objects from a set may be selected, generally without replacement to from subset. It is denoted as ${}^{n}{{P}_{k}}$ where n is the total number of objects and k means how many to get selected. Using this concept, we will answer.

Complete step by step answer:
In the question, we are asked to find How many 6 letter words can be formed by taking letters of the word EDUCATION.
So, there are a total 9 letters in the word EDUCATION and no letter is repeated in that word. So, we have total 6 blanks with us i.e. $-,-,-,-,-,-$
Now, let us assume that out of 9 letters i.e. E, D, U, C, A, T, I, O, N There is any letter in the first blank. So, we can write it as ${}^{9}{{P}_{6}}$ because we have to do a selection of 6 letters only.
Now, next we have to find how many words can be formed with the letter I am starting. So, out of all the 6 blanks, first blank will always be with letter I i.e. $I,-,-,-,-,-$ . Thus, we have now total 8 letters remaining and 5 blanks. So, here the selection will be like out of a total 8 letter, only 5 to be selected written as ${}^{8}{{P}_{5}}$ .
Thus, ${}^{9}{{P}_{6}}$ 6-letter words with distinct letters in each can be formed using the letters of the word EDUCATION and ${}^{8}{{P}_{5}}$ words of these begin with I.
Hence, option (a) is the correct answer.

Note: Sometimes mistake happens in considering that words starting with letter I. Students assume the blanks as first blank will always be with letter I i.e. $I,-,-,-,-,-$ . Now, we have to subtract one letter out of the total 9 letters, so remaining letters will be 8. But this step students missed in hurry and the answer results out of 9 letters only we have to select 5 remaining blanks. So, it will be ${}^{9}{{P}_{5}}$ . But this answer is wrong. So, do not make this mistake.